Probability, Permutation and Combination

PrepBank · পাতা ১০ / ১০ · ৯০১–৯৬৫ / ৯৬৯

৯০১.

There are 8 marbles in a box 6 red and 2 black. If you randomly pick 2 marbles simultaneously what is the probability that you will get one red and 1 black marble?

ক
ক) 3/7
খ
খ) 3/14
গ
গ) 3/8
ঘ
ঘ) None

ব্যাখ্যা

Question: There are 8 marbles in a box - 6 red and 2 black. If you randomly pick 2 marbles simultaneously what is the probability that you will get one red and 1 black marble?

Solution:
লাল বল = 6 টি
কালো বল = 2টি
মোট বল = (6 + 2)টি = 8টি

6টি লাল বল থেকে 1টি লাল বল আসবে =⁶C₁
2টি কালো বল থেকে 1টি কালো বল আসবে = ²C₁

8টি বল থেকে 2টি বল আসবে = ⁸C₂

1টি লাল ও 1টি কালো বল হওয়ার সম্ভাবনা = (⁶C₁ × ²C₁)/⁸C₂
= (6 × 2)/28
= 3/7

৯০২.

A card is randomly drawn from a deck of 52 cards. What is the probability of getting a king or Queen?

ক
ক) 3/13
খ
খ) 2/13
গ
গ) 1/13
ঘ
ঘ) 4/13

ব্যাখ্যা

Total number of kings is 4 out of 52 cards.
Total number of queens is 4 out of 52 cards
Number of favorable outcomes i.e. ‘a king or a queen’ is 4 + 4 = 8 out of 52 cards.
Therefore, probability of getting ‘a king or a queen’ = Number of favorable outcomes
∴ P(E) = Total number of possible outcomes = 8/52 = 2/13

৯০৩.

In how many ways can a leap year have 53 Mondays?

ক
8
খ
6
গ
4
ঘ
2

ব্যাখ্যা

Question: In how many ways can a leap year have 53 Mondays?

Solution:
In a leap year there are 366 days
52 weeks + 2 extra days.
So to have 53 Mondays one of these two days must be a Monday.
This can occur in only 2 ways.
(Sunday and Monday) or (Monday and Tuesday).
Thus number of ways = 2

৯০৪.

Three coins are tossed. Find the probability of exactly 2 heads -

ক
(3/8)
খ
(1/2)
গ
(1/8)
ঘ
None

ব্যাখ্যা

Question: Three coins are tossed. Find the probability of exactly 2 heads -

Solution:
n(S) = 2³ = 8
Let E = Event of getting exactly two heads,
= {(H, H, T), (H, T, H), (T, H, H)}
= n(E)
= 3

Required probability = 3/8

৯০৫.

A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If four balls are picked up at random, what is the probability that 2 are red and 2 are green?

ক
5/197
খ
3/293
গ
2/455
ঘ
4/185
ঙ
None

ব্যাখ্যা

Question: A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If four balls are picked up at random, what is the probability that 2 are red and 2 are green?

Solution:
Total number of balls = 6 + 2 + 4 + 3 = 15.
Let E be the event of drawing 4 balls such that 2 are red and 2 are green.
Then, n(E) = (²C² × ⁴C₂) = 6
And n(S) = ¹⁵C₄ = 365

∴ P(E) = n(E)/n(S)
= 6/1365
= 2/455

৯০৬.

In a box, there are 4 red, 6 white, and 2 black balls. If two balls are drawn one after the other without replacement, what is the probability that the first ball is red and the second ball is white?

ক
2/11
খ
3/7
গ
5/13
ঘ
2/13
ঙ
None of these

ব্যাখ্যা

Question: In a box, there are 4 red, 6 white, and 2 black balls. If two balls are drawn one after the other without replacement, what is the probability that the first ball is red and the second ball is white?

Solution:
মোট বলের সংখ্যা = 4 (লাল) + 6 (সাদা) + 2 (কালো) = 12টি।
∴ প্রথম বলটি লাল হওয়ার সম্ভাবনা = 4/12 = 1/3

যেহেতু বলটি প্রতিস্থাপন করা হয়নি (Without replacement),
তাই প্রথম বলটি তোলার পর বক্সে মোট বলের সংখ্যা = 12 - 1 = 11টি।
∴ দ্বিতীয় বলটি সাদা হওয়ার সম্ভাবনা = 6/11

∴ প্রথমটি লাল এবং দ্বিতীয়টি সাদা হওয়ার সম্ভাবনা = (প্রথমটি লাল হওয়ার সম্ভাবনা) × (দ্বিতীয়টি সাদা হওয়ার সম্ভাবনা)
= (1/3) × (6/11)
= 6/33
= 2/11

∴ নির্ণেয় সম্ভাবনা 2/11

৯০৭.

All possible three-digit numbers are formed by the digits 1, 2, 3, 5, 6 (without repetition). If one number is chosen randomly, what is the probability that it is divisible by 5?

ক
1/3
খ
2/5
গ
5/12
ঘ
1/5

ব্যাখ্যা

Question: All possible three-digit numbers are formed by the digits 1, 2, 3, 5, 6 (without repetition). If one number is chosen randomly, what is the probability that it is divisible by 5?

Solution:
প্রদত্ত অঙ্কগুলো: 1, 2, 3, 5, 6

তাহলে, তিন অঙ্কের মোট সংখ্যা = ⁵P₃ = 5 × 4 × 3 = 60 টি।

একটি সংখ্যা 5 দ্বারা বিভাজ্য হবে যদি তার শেষ অংকটি 5 হয়।
• শেষ অংকটি 5 (১টি উপায়)।
• বাকি চারটি অংক (1, 2, 3, 6) থেকে প্রথম দুটো স্থান পূরণ করতে হবে।
• প্রথম দুটো স্থান পূরণ করার উপায় = ⁴P₂ = 4 × 3 = 12 টি।

সুতরাং, 5 দ্বারা বিভাজ্য মোট সংখ্যা = 12 টি।

অতএব, সংখ্যাটি 5 দ্বারা বিভাজ্য হবার সম্ভাবনা = (অনুকূল ফলাফল)/(মোট ফলাফল)
= 12/60
= 1/5

৯০৮.

In how many ways can a group of 4 men and 3 women be made out of a total of 6 men and 5 women?

ক
120 ways
খ
150 ways
গ
210 ways
ঘ
90 ways

ব্যাখ্যা

Question: In how many ways can a group of 4 men and 3 women be made out of a total of 6 men and 5 women?

Solution:
There are 6 men and 5 women. We have to select 4 men out of 6 and 3 women out of 5.

Number of ways to select 4 men from 6 = ⁶C₄ = 6!/(4! × 2!)
= (6 × 5)/(2 × 1) = 15

Number of ways to select 3 women from 5 = ⁵C₃
= 5!/(3! × 2!)
= (5 × 4)/(2 × 1)
= 10

∴ The number of ways of making the selection = 15 × 10 = 150 ways

৯০৯.

A bag contains 30 balls numbered 1 to 30. Two balls are drawn at random. What is the probability that the balls drawn contain a number which is multiple of 4 or 6 but not a multiple of both?

ক
ক) 1/8
খ
খ) 1/4
গ
গ) 1/3
ঘ
ঘ) None

ব্যাখ্যা

Question: A bag contains 30 balls numbered 1 to 30. Two balls are drawn at random. What is the probability that the balls drawn contain a number which is multiple of 4 or 6 but not a multiple of both?

Solution:
Total outcomes are 30
LCM of 4 and 6 is 12.
Common numbers are 30/12 ≈ 2
Now, favourable outcomes are (28/4) - 2 + (30/6) - 2 = 8

So, Probability is (8/30) × (7/29) = 28/435

৯১০.

One ball is picked up randomly from a bag containing 8 yellow, 7 blue and 6 black balls. What is the probability that it is neither yellow nor black?

ক
1/2
খ
1/3
গ
1/4
ঘ
3/4
ঙ
3/5

ব্যাখ্যা

Total number of balls, n(S) = 8 + 7 + 6 = 21
n(E) = Number of ways in which a ball can be selected which is neither yellow nor black = 7 (∵ there are only 7 balls which are neither yellow nor black)
P(E) = n(E)/n(S) = 7/21 = 1/3

৯১১.

In how many different ways can the letters of the word INCREASE be arranged?

ক
ক) 20016
খ
খ) 20160
গ
গ) 16200
ঘ
ঘ) 22016

ব্যাখ্যা

The given words contains 8 letters of which E is taken 2 times.
∴ Required number of ways = 8!/2!
= (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/2
= 20160

৯১২.

How many triangles can be formed with 18 points?

ক
816
খ
720
গ
1024
ঘ
924

ব্যাখ্যা

Question: How many triangles can be formed with 18 points?

Solution:
আমরা জানি,
একটি ত্রিভুজ গঠন করতে ৩ টি বিন্দু প্রয়োজন হয়।
তাহলে,
18 টি বিন্দু দিয়ে গঠিত ত্রিভুজের সংখ্যা = ¹⁸C₃
= 18!/{3! × (18 - 3)!}
= 18!/(3! × 15!)
= (18 × 17 × 16 × 15!)/(3 × 2 × 1 × 15!)
= (18 × 17 × 16)/(3 × 2 × 1)
= (18 × 17 × 16)/6
= 4896/6
= 816

∴ 18 টি বিন্দু দিয়ে মোট 816 টি ত্রিভুজ গঠন করা যাবে।

৯১৩.

Two dice are rolled. What is the probability that the sum is 7?

ক
1/6
খ
1/12
গ
5/24
ঘ
7/36

ব্যাখ্যা

Question: Two dice are rolled. What is the probability that the sum is 7?

Solution:
When two dice are rolled, the total number of possible outcomes is = 6 × 6 = 36
Favorable Outcomes (Sum = 7) = 6 [(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)]

Probability = 6/36 = 1/6

৯১৪.

If ⁶P_r = 120, what is the value of r?

ক
2
খ
3
গ
4
ঘ
12

ব্যাখ্যা

Question: If ⁶P_r = 120, what is the value of r?

Solution:
আমরা জানি, ⁿP_r = n!/(n - r)!
দেওয়া আছে,
⁶P_r = 120
⇒ 6!/(6 - r)! = 120
⇒ 720/(6 - r)! = 120 (কারণ 6! = 720)
⇒ (6 - r)! = 720/120
⇒ (6 - r)! = 6
⇒ (6 - r)! = 3! (কারণ 3! = 3 × 2 × 1 = 6)
⇒ 6 - r = 3
⇒ r = 6 - 3
∴ r = 3

৯১৫.

A box contains three green balls, four white balls, and three red balls. The number of ways in which three balls can be drawn from the box so that at least one of the balls is white is-

ক
60
খ
80
গ
100
ঘ
120

ব্যাখ্যা

Question: A box contains three green balls, four white balls, and three red balls. The number of ways in which three balls can be drawn from the box so that at least one of the balls is white is-

Solution:
The required number of ways
1 white and 2 others = ⁴C₁ × ⁶C₂ = 4 × 15 = 60
2 white and 1 other = ⁴C₂ × ⁶C₁ = 6 × 6 = 36
All the three white = ⁴C₃ = 4

∴ Total = 60 + 36 + 4 = 100

৯১৬.

Four boys and four girls are to be seated alternately around a round table. In how many different ways can this be done?

ক
192
খ
288
গ
144
ঘ
720

ব্যাখ্যা

Question: Four boys and four girls are to be seated alternately around a round table. In how many different ways can this be done?

Solution:

Let’s fix 1 boy in one seat (since it's a round table).

Remaining boys = 4 - 1 = 3
The number of ways to arrange the remaining 3 boys = 3! = 6
The number of ways to arrange the 4 girls is 4! = 24

Total arrangements = 6 × 24 = 144

৯১৭.

A committee is to consist of three members. If there are seven men and five women available to serve on the committee, how many different committees can be formed?

ক
ক) 1320
খ
খ) 350
গ
গ) 220
ঘ
ঘ) 120

ব্যাখ্যা

Question: A committee is to consist of three members. If there are seven men and five women available to serve on the committee, how many different committees can be formed?

Solution:
Total, n = 7 + 5 = 12
and r = 3

Number of committee = ¹²C₃
= 220

৯১৮.

If the probability of rain on any given day in City Dhaka is 50 percent, what is the probability that it rains on exactly 2 days in a 4-day period?

ক
ক) 1/16
খ
খ) 1/2
গ
গ) 3/8
ঘ
ঘ) None of these

ব্যাখ্যা

Question: If the probability of rain on any given day in City Dhaka is 50 percent, what is the probability that it rains on exactly 2 days in a 4-day period?

Solution:
If the probability of rain on any given day in City Dhaka is 50 percent
the probability of rain on any given day = 1/2
the probability of no rain on any given day = 1/2

selecting 2 days out of 4 = ⁴C₂

∴the probability that it rains on exactly 2 days in a 4-day period is = ⁴C₂ × 1/2 × 1/2 × 1/2 × 1/2
= 6 × 1/2⁴
= 6 × 1/16
= 3/8

৯১৯.

In how many different ways can the letters of the word ‘BALLOON’ be arranged?

ক
1260
খ
1060
গ
1660
ঘ
1560

ব্যাখ্যা

Question: In how many different ways can the letters of the word ‘BALLOON’ be arranged?

Solution:
Number of letters in the word = 7

Repeated letters:
L = 2 times
O = 2 times
The remaining letters (B, A, N) are different.

∴ Number of arrangements
= 7!/(2! × 2!)
= 5040/4
= 1260

৯২০.

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

ক
1/2
খ
3/4
গ
3/8
ঘ
5/16

ব্যাখ্যা

Question: Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

Solution:
In a simultaneous throw of two dice, we have n(S) = (6 × 6) = 36

Then,
E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(E) = 27

∴ P(E) = n(E)/n(S) = 27/36 = 3/4

৯২১.

⁴P₂ - ⁴C₂ =?

ক
12
খ
8
গ
6
ঘ
4

ব্যাখ্যা

প্রশ্ন: ⁴P₂ - ⁴C₂ =?

সমাধান:
⁴P₂
= 4!/(4 - 2)!
= 4!/2!
= 12

⁴C₂
= 4!/2!(4 - 2)!
= 4!/2! 2!
= 6

∴ ⁵P₂ - ⁵C₂ = 12 - 6
= 6

৯২২.

In a party every person shakes hands with every other person. If there are 78 hands shakes, find the number of person in the party.

ক
12
খ
13
গ
14
ঘ
15

ব্যাখ্যা

Question: In a party every person shakes hands with every other person. If there are 78 hands shakes, find the number of person in the party.

Solution:
Let n be the number of persons in the party
Number of hands shake = 78
Total number of hands shake is given by = ⁿC₂
Now,
According to the question,
ⁿC₂= 78
⇒ n!/{2!(n - 2)!} = 78
⇒ {n × (n - 1)}/2 = 78
⇒ n^{2 _-}n =156
⇒ n²- 13n + 12n - 156 = 0
⇒ n(n - 13) + 12(n - 13) = 0
or, n = 13, -12

But, we cannot take negative value of n
So, n = 13
∴ number of persons in the party = 13

৯২৩.

X = { - 4, - 2, 1, 3}, Y = {- 1, 4, 5}. If x is a number from set X, and y is a number from set Y. The probability that x + y is positive is closest to-

ক
ক) 0.5
খ
খ) 0.6
গ
গ) 0.7
ঘ
ঘ) 0.8

ব্যাখ্যা

Question: X = { - 4, - 2, 1, 3}, Y = {- 1, 4, 5}. If x is a number from set X, and y is a number from set Y. The probability that x + y is positive is closest to-

Solution:
X = { - 4, - 2, 1, 3}
Y = {- 1, 4, 5}

X সেটের উপাদান সংখ্যা = 4
Y সেটের উপাদান সংখ্যা = 3

মোট ফলাফল = ⁴C₁ × ³C₁ = 4 × 3 = 12
অনুকূল ঘটনার সংখ্যা = {( - 4, 5), ( - 2, 4), ( - 2, 5), (1, 4), (1, 5), (3,-1), (3, 4), (3, 5)}
অনুকূল ঘটনার সংখ্যা = 8

নির্ণেয় সম্ভাব্যতা = 8/12
= 2/3
= 0.666
≈ 0.7

৯২৪.

Three unbiased coins are tossed. What is the probability of getting at least 2 heads?

ক
1/4
খ
2/3
গ
1/2
ঘ
1/3

ব্যাখ্যা

Question: Three unbiased coins are tossed. What is the probability of getting at least 2 heads?

Solution:
Total outcomes = {TTT, TTH,THT, HTT, THH, HTH, HHT, HHH} = 8
Favorable outcomes = {HHT, HTH, THH, HHH} = 4

So, the probability of getting at least 2 heads = Favorable outcomes/Total outcomes
= 4/8 = 1/2

৯২৫.

In a simultaneous throw of two dice what is the probability of getting a total of 10 or 11?

ক
ক) 7/12
খ
খ) 1/6
গ
গ) 5/36
ঘ
ঘ) 1/4

ব্যাখ্যা

Question: In a simultaneous throw of two dice what is the probability of getting a total of 10 or 11?

Solution:
If two dices are thrown total events = (6 × 6) = 36
Event of getting a total of 10 or 11 = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)}
Expected events = 5

∴ Probability = 5/36

৯২৬.

An unbiased die is tossed.Find the probability of getting a multiple of 3.

ক
ক) 1/3
খ
খ) 1/2
গ
গ) 3/4
ঘ
ঘ) 3/2
ঙ
ঙ) None of the Above

ব্যাখ্যা

Here S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting the multiple of 3
Then,
E = {3,6}
P(E) = n(E)/n(S)
= 2/6
= 1/3

৯২৭.

If 5 × ⁿP₃ = 4 × ^{(n + 1)}P₃, find n?

ক
ক) 10
খ
খ) 12
গ
গ) 13
ঘ
ঘ) 14

ব্যাখ্যা

Question: If 5 × ⁿP₃ = 4 × ^{(n + 1)}P₃, find n?

Solution:
5 × ⁿP₃= 4 × ^{(n + 1)}P₃
5 × n × (n - 1) × (n - 2) = 4 × (n + 1) × n × (n - 1)
Or, 5(n - 2) = 4(n + 1)
Or, 5n - 10 = 4n + 4
Or, 5n - 4n = 4 + 10
Hence, n = 14

৯২৮.

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7?

ক
ক) 1/20
খ
খ) 4/9
গ
গ) 2/5
ঘ
ঘ) 4/5

ব্যাখ্যা

Question: Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7?

Solution:
multiple of 3 = 3, 6, 9, 12, 15, 18
multiple of 7 = 7, 14

So, probability that the ticket drawn has a number which is a multiple of 3 or 7
= probability of getting multiple 3 + probability of getting multiple 7
= (6/20) + (2/20)
= (3/10) + (1/10)
= 4/10
= 2/5

৯২৯.

A batsman hits boundaries 6 times out of 30 balls. Find the probability that he did not hit the boundaries.

ক
1/5
খ
2/5
গ
3/5
ঘ
4/5
ঙ
2/3

ব্যাখ্যা

No. of boundaries = 6
No. of balls = 30
No. of balls without boundaries = 30 – 6 =24
Probability of no boundary = 24/30 = 4/5

৯৩০.

In how many ways can 5 boys and 4 girls be seated in a row so that they sit alternately?

ক
1880
খ
2880
গ
1550
ঘ
1200

ব্যাখ্যা

Question: In how many ways can 5 boys and 4 girls be seated in a row so that they sit alternately?

Solution:
Let the arrangement be,
B G B G B G B G B

5 boys can be seated in 5! ways
4 girls can be seated in 4! ways

∴ Required number of ways
= 5! × 4!
= 120 × 24
= 2880

৯৩১.

Q. (33-56): Choose the correct answer.
In a container, there are 2 green marbles and 2 red marbles. You randomly pick the marbles. What is the probability that both of them are green?

ক
ক) 1/2
খ
খ) 1/3
গ
গ) 1/4
ঘ
ঘ) 1/6

ব্যাখ্যা

Question: In a container, there are 2 green marbles and 2 red marbles. You randomly pick the marbles. What is the probability that both of them are green?

Solution:
দুইটি সবুজ ও দুইটি লাল মার্বেল।
মোট মার্বেল = ৪ টি
এর মধ্যে একটি তুললে সবুজ হওয়ার সম্ভাবনা = ২/৪ = ১/২

একটি তুলার পর মার্বেল থাকে ৩ টি যার মধ্যে সবুজ আছে ১ টি
এই তিনটির মধ্যে একটি মার্বেল তুললে এটি সবুজ হওয়ার সম্ভাবনা = ১/৩

তাহলে দুইটি মার্বেল সবুজ হওয়ার সম্ভাবনা = (১/২)(১/৩) = ১/৬

৯৩২.

In how many ways can 5 balls can be chosen from 9 different balls?

ক
102
খ
110
গ
118
ঘ
126

ব্যাখ্যা

Question: In how many ways can 5 balls can be chosen from 9 different balls?

Solution:

Here,
Total number of different balls, n = 9
Chosen balls from different balls, r = 5

The number of ways 5 balls can be chosen is
= ⁿC_r=n!/r!(n - r)!
= 9!/5!(9 - 5)!
= 9!/(5! × 4!)
= (9 × 8 × 7 × 6 × 5!)/(5! × 4!)
= (9 × 8 × 7 × 6)/4!
= 3024/(4 × 3 × 2 × 1)
= 3024/24
= 126

∴ 5 balls can be chosen from 9 different balls in 126 ways.

৯৩৩.

In a box, there are 5 red, 9 blue, and 10 green balls. One ball is picked randomly. What is the probability that it is neither red nor green?

ক
5/8
খ
3/8
গ
2/5
ঘ
3/5

ব্যাখ্যা

Question: In a box, there are 5 red, 9 blue, and 10 green balls. One ball is picked randomly. What is the probability that it is neither red nor green?

Solution:
Total number of balls = 5 + 9 + 10 = 24

Let, E be the event that the ball is neither red nor green = the event that the ball is blue.
∴ n(E) = 9

∴ Probability = Number of favorable outcomes/Total outcomes
= 9/24
= 3/8

৯৩৪.

There are 35 people in a room. If each person shakes hands with every other person exactly once, how many handshakes will take place in total?

ক
295
খ
495
গ
595
ঘ
195
ঙ
None

ব্যাখ্যা

Question: There are 35 people in a room. If each person shakes hands with every other person exactly once, how many handshakes will take place in total?

Solution:
Total handshakes = ³⁵C₂
= (35 × 34)/2
= 595

∴ Total handshakes = 595

৯৩৫.

What is the probability that an integer selected at random from those between 20 and 100 inclusive is a multiple of 15?

ক
5/81
খ
3/78
গ
7/79
ঘ
4/83

ব্যাখ্যা

Multiple of 15 from 20 to 100 is 30, 45, 60, 75, 90 = 5
∴ Probability = 5/81

৯৩৬.

There are 4 women and 4 men sitting in a waiting room for a job interview. If two of the applicants are selected at random, what is the probability that both will be women?

ক
ক) 1/2
খ
খ) 3/7
গ
গ) 3/4
ঘ
ঘ) 3/14

ব্যাখ্যা

৮ জনের মধ্যে ৪ জন মহিলা ও ৪ জন পুরুষ।
দুইজনকে দৈবভাবে নিলে মহিলা আসার সম্ভাবনা বের করতে হবে।
প্রথম ১ জন নিলে মহিলা আসার সম্ভাবনা = মোট মহিলা/মোট সংখ্যা
= ৪/৮
= ১/২
তখন মোট সংখ্যা থাকবে ৭ জন ও মহিলা থাকবে ৩ জন।
তারপর ১ জন নিলে মহিলা আসার সম্ভাবনা = ৩/৭।
তাহলে দুইজন নিলে একত্রে সম্ভাবনা হবে = ১/২ × ৩/৭
= ৩/১৪।

৯৩৭.

Wendy has 5 pairs and 8 shirts. How many different combinations can she make with these items?

ক
42
খ
40
গ
38
ঘ
36

ব্যাখ্যা

Question: Wendy has 5 pairs and 8 shirts. How many different combinations can she make with these items?

Solution:
Wendy can make = 5 × 8 = 40 different combinations.

৯৩৮.

A box contains 21 balls numbered 1 to 21 . A ball is drawn and then another ball is drawn without replacement. What is the probability that both balls are even numbered?

ক
ক) 2/7
খ
খ) 2/21
গ
গ) 3/14
ঘ
ঘ) 10/21

ব্যাখ্যা

Question: A box contains 21 balls numbered 1 to 21 . A ball is drawn and then another ball is drawn without replacement. What is the probability that both balls are even numbered?

Solution:
The probability that first ball shows the even number =10/ 21

Since, the ball is not replaced there are now 9 even numbered balls and total 20 balls left.
Hence, probability that second ball shows the even number =9/20

Required probability,
=(10/21) × (9/20) = 3/14

৯৩৯.

6 people at a party shake hands once with everyone else in the room.How many handshakes took place?

ক
ক) 10
খ
খ) 12
গ
গ) 15
ঘ
ঘ) 24

ব্যাখ্যা

Question: 6 people at a party shake hands once with everyone else in the room.How many handshakes took place?

Solution:
handshakes took place = ⁶C₂
= 6!/2! 4!
= 15

৯৪০.

In how many different ways can the letters of the word "ACTIVE" be arranged so that the vowels occupy only odd positions?

ক
24
খ
30
গ
36
ঘ
18

ব্যাখ্যা

Question: In how many different ways can the letters of the word "ACTIVE" be arranged so that the vowels occupy only odd positions?

Solution:
The word "ACTIVE" consists of the letters, Vowels are (A, I, E) and Consonants (C, T, V)

The number of ways to arrange the 3 vowels in the 3 odd positions is = 3!
= 6

The number of ways to arrange the 3 consonants in the 3 even positions is = 3! = 6

∴ The requried ways = (6 × 6)
= 36

৯৪১.

In how may different ways can the letters of the word 'PIRATE' be arranged in such a way that the vowels always come together?

ক
ক) 720
খ
খ) 360
গ
গ) 144
ঘ
ঘ) None of the above

ব্যাখ্যা

Question: In how may different ways can the letters of the word PIRATE be arranged in such a way that the vowels always come together?

Solution:
The word 'PIRATE' contains 6 different letters.
When the vowels IAE are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PRT(IAE).
Now, 4 letters can be arranged in 4! ways
= 24 ways.
The vowels (IAE) can be arranged among themselves in 3! = 6 ways.

∴ Required number of ways = (24 x 6) = 144

৯৪২.

There are 7 non-collinear points. How many triangles can be drawn by joining these points?

ক
28
খ
35
গ
46
ঘ
54

ব্যাখ্যা

Question: There are 7 non-collinear points. How many triangles can be drawn by joining these points?

Solution:
A triangle is formed by joining any three non-collinear points in pairs.
There are 7 non-collinear points.

∴ The number of triangles formed = ⁷C₃ = 35

৯৪৩.

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

ক
ক) 1/9
খ
খ) 5/9
গ
গ) 8/9
ঘ
ঘ) 4/9

ব্যাখ্যা

One person can select one house out of 3 = ³C₁ ways =3 ways
Hence, three persons can select one house in = 3 × 3 × 3 = 27 ways
So, All three apply for the same house, its probability = 3/27 = 1/9

৯৪৪.

A committee of 5 members is to be made of 5 males and 10 females so that there is always 3 females on the committee. How many ways can the committee be formed?

ক
600
খ
720
গ
480
ঘ
1200
ঙ
2400

ব্যাখ্যা

Question: A committee of 5 members is to be made of 5 males and 10 females so that there is always 3 females on the committee. How many ways can the committee be formed?

Solution:
As there should be always 3 females in the committee,
total ways = ⁵C₂ × ¹⁰C₃
= 10 × 120
= 1200

৯৪৫.

A college has 10 basketball players. A 5 member's team and a captain will be selected out of these 10 players. How many different selections can be made?

ক
1260
খ
1180
গ
190
ঘ
1140

ব্যাখ্যা

Question: A college has 10 basketball players. A 5 member's team and a captain will be selected out of these 10 players. How many different selections can be made?

Solution:
We can select the 5 member team out of the 10 in = ¹⁰C₅ ways
= 252 ways

The captain can be selected from amongst the remaining 5 players in 5 ways.

∴ The total ways the selection of 5 players and a captain can be made = 252 × 5 ways
= 1260 ways

৯৪৬.

Numbers from 10 to 20 are written on cards. One card is drawn at random. What is the probability that the number is prime or divisible by 3?

ক
5/11
খ
6/11
গ
8/11
ঘ
7/11

ব্যাখ্যা

Question: Numbers from 10 to 20 are written on cards. One card is drawn at random. What is the probability that the number is prime or divisible by 3?

Solution:
The numbers are,
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
∴ Total numbers = 11

∴ Prime number are, 11, 13, 17, 19 = 4

And, List the numbers divisible by 3 = 3 numbers (12, 15, 18)

∴ Total favorable outcomes = Primes + Divisible by 3 - Overlap
= 4 + 3 - 0 = 7

∴ Probability = Favorable outcomes/Total outcomes
= 7/11

So the probability is 7/11.

৯৪৭.

Six friends Rina, Toma, Jui, Lima, Arif, and Sayeed sit randomly in a row of six chairs. What is the probability that Rina and Toma do not sit next to each other?

ক
1/2
খ
1/3
গ
2/3
ঘ
5/6
ঙ
None

ব্যাখ্যা

Question: Six friends Rina, Toma, Jui, Lima, Arif, and Sayeed sit randomly in a row of six chairs. What is the probability that Rina and Toma do not sit next to each other?

Solution:
Total number of possibilities = 6! = 720

Number of possibilities where Rina and Toma sit together = 5! × 2!
= 120 × 2
= 240

So the possibilities where Rina and Toma do not sit together = 720 - 240
= 480

∴Probability that Rina and Toma do not sit next to each other = 480/720
= 2/3

৯৪৮.

In how many different ways can the letters of the word 'DETAIL' be arranged so that the vowels occupy only the odd positions?

ক
18
খ
26
গ
36
ঘ
44
ঙ
58

ব্যাখ্যা

Question: In how many different ways can the letters of the word 'DETAIL' be arranged so that the vowels occupy only the odd positions?
('DETAIL' শব্দের অক্ষরগুলি কতভাবে সাজানো সম্ভব, যেখানে স্বরবর্ণগুলো শুধুমাত্র বিজোড় স্থানেই থাকবে?)

Solution:
there are 6 letters, where there are 3 vowels and 3 consonants.

3 vowels in 3 odd positions can be arranged in = ³P₃
= 3! = 6 ways
3 consonants in 3 even positions can be arranged in = ³P₃
= 3! = 6 ways

total ways = 6 × 6 = 36 ways

৯৪৯.

How many different ways can the letters of the word "LEVEL" be arranged?

ক
25
খ
40
গ
120
ঘ
30

ব্যাখ্যা

Question: How many different ways can the letters of the word "LEVEL" be arranged?

Solution:
The given word contains 5 letters, where L and E is taken 2 times.

∴ Required number of ways = 5!/(2! × 2!) = 120/4 = 30

৯৫০.

In how many ways can 4 students be chosen from a class of 12 students?

ক
525
খ
554
গ
582
ঘ
590
ঙ
None of the above

ব্যাখ্যা

Question: In how many ways can 4 students be chosen from a class of 12 students?

Solution:
Here, total number of students, n = 12
Number of students to be chosen, r = 4
We know, the number of ways to choose r objects from n objects is ⁿC_r
So, total ways = ¹²C₄
= 12!/{4! × (12 - 4)!}
= 12!/(4! × 8!)
= (12 × 11 × 10 × 9 × 8!)/(4 × 3 × 2 × 1 × 8!)
= (12 × 11 × 10 × 9)/24
= 11880/24
= 495

৯৫১.

If 10 person shake their hands with each other , then total number of handshakes are-

ক
100
খ
50
গ
45
ঘ
35

ব্যাখ্যা

Question: If 10 person shake their hands with each other , then total number of handshakes are-

Solution:
আমরা জানি,
পরস্পর হ্যান্ডশেক করতে 2 জন ব্যাক্তির প্রয়োজন হয়।

∴ 10 জন লোক যদি পরস্পর পরস্পরের সাথে হ্যান্ডশেক করে, তাহলে মোট হ্যান্ডশেক এর সংখ্যা হবে,
= ¹⁰C₂
= 10!/{2! × (10-2)!}
= 10!/(2! × 8!)
= (10 × 9 × 8!)/(2! × 8!))
= (10 × 9)/(2 × 1)
= 45

৯৫২.

A box contains three colored marbles, such that the number of blue, yellow and grey marbles is 8, 7 and 5 respectively. Find the probability of picking two marbles, one being blue marble and the other being grey marble.

ক
3/19
খ
5/19
গ
2/19
ঘ
4/19

ব্যাখ্যা

Question: A box contains three colored marbles, such that the number of blue, yellow and grey marbles is 8, 7 and 5 respectively. Find the probability of picking two marbles, one being blue marble and the other being grey marble.

Solution:
From the given data,
Number of blue marbles = 8
Number of yellow marbles = 7
Number of grey marbles = 5

∴ Probability = (⁸C₁ × ⁵C₁)/²⁰C₂
= (8 × 5)/190
= 40/190
= 4/19.
Hence, option(4) is correct.

৯৫৩.

At a conference, everyone shakes hands with everybody else. If there were 105 handshakes, how many people were at the conference?

ক
12
খ
14
গ
15
ঘ
19

ব্যাখ্যা

Question: At a conference, everyone shakes hands with everybody else. If there were 105 handshakes, how many people were at the conference?

Solution:
ধরি, কনফারেন্সে x সংখ্যক লোক ছিল।
দুজন ব্যক্তির মধ্যে একটি হ্যান্ডশেক হয়, যা Combination দিয়ে প্রকাশ করা যায়।

প্রশ্নমতে,
^xC₂ = 105
⇒ x!/(2!(x - 2)!) = 105
⇒ {x(x - 1)(x - 2)!}/{2 × 1 × (x - 2)!} = 105
⇒ x(x - 1)/2 = 105
⇒ x(x - 1) = 210
⇒ x² - x - 210 = 0
⇒ x² - 15x + 14x - 210 = 0
⇒ x(x - 15) + 14(x - 15) = 0
⇒ (x - 15)(x + 14) = 0

হয় x - 15 = 0 অথবা x + 14 = 0
∴ x = 15 অথবা x = -14

যেহেতু লোকসংখ্যা ঋণাত্মক হতে পারে না, তাই x = 15

অতএব, কনফারেন্সে 15 জন লোক ছিল।

৯৫৪.

What is the probability of getting a sum of six if two dice are thrown at once?

ক
ক) 5/36
খ
খ) 7/36
গ
গ) 2/3
ঘ
ঘ) 1/36

ব্যাখ্যা

We know, the probability of an event = Favorable outcomes / Total outcomes.
In a case of two dices, Total outcomes = 6 × 6 = 36
A.T.Q favorable outcome is a sum of 6,
Ways of obtaining a sum of 6 (1,5), (5,1), (2,4), (4,2) and (3,3).
Thus, there are 5 ways in which 6 can be obtained using two dices.
Therefore, required probability, P = 5/36

৯৫৫.

A coin is tossed four times, if H = head and T = tail, what is the probability of the tosses coming up in the order HHTH?

ক
1/8
খ
3/8
গ
1/16
ঘ
1/4

ব্যাখ্যা

Question: A coin is tossed four times, if H = head and T = tail, what is the probability of the tosses coming up in the order HHTH?

Solution:
Here, Coin is tossed four times.
The total possible outcomes = 16
Outcome = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

Favourable outcome = 1 [HHTH]
∴ Probability = 1/16

৯৫৬.

There are 3 green, 4 orange and 5 white color bulbs. If a bulb is picked at random, what is the probability of having either a green or a white bulb?

ক
ক) 2/3
খ
খ) 3/4
গ
গ) 4/3
ঘ
ঘ) 2/5

ব্যাখ্যা

সবুজ রঙের বাল্ব = 3
কমলা রঙের বাল্ব = 4
সাদা রঙের বাল্ব =5

মোট বাল্ব আছে = 3 + 4 + 5 = 12 টি

মোট সবুজ ও সাদা বাল্ব আছে = (3 + 5) = 8টি
একটি সবুজ বা সাদা বাল্ব পাওয়ার সম্ভাবনা = 8/12 = 2/3

৯৫৭.

What is the probability of getting a sum of 10 from two throws of a dice?

ক
2/5
খ
3/10
গ
1/12
ঘ
7/10

ব্যাখ্যা

Question: What is the probability of getting a sum of 10 from two throws of a dice?

Solution:
Total events = (6 × 6) = 36
events to get a sum of 10 is -
{(6, 4), (5, 5), (4, 6)}

probability = 3/36 = 1/12

৯৫৮.

In how many different ways can the letters of the word ''CANDIDATE'' be arranged in such a way that the vowels always come together?

ক
4320
খ
1440
গ
720
ঘ
840

ব্যাখ্যা

There are 9 letters in the given word, out of which 4 are vowels.
In the word ''CANDIDATE'' we treat the vowels ''AIAE'' as one letter.
Thus, we have CNDDT(AIAE).
Now, we have to arrange 6 letters, out of which D occurs twice.
Therefore,
Number of ways of arranging these letters = 6!/2!
= 720/2
= 360 ways.
Now,
AIAE has 4 letters, in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4!/2! = (1 × 2 × 3 × 4)/2
= 12
Therefore, required number of words = (360 × 12)
= 4320.

৯৫৯.

A die is thrown three times and the sum of the three numbers is found to be 15. The probability that the first throw was a four is

ক
1/108
খ
2/9
গ
1/5
ঘ
1/2

ব্যাখ্যা

Question: A die is thrown three times and the sum of the three numbers is found to be 15. The probability that the first throw was a four is

Solution:
6 + 6 + 3 = 15, number of ways = 3!/2! = 3
6 + 5 + 4 = 15, number of ways = 3! = 6
5 + 5 +5 = 15, number of ways = 1
Total ways = 10

If first throw was a four = (4, 5, 6), (4, 6, 5) = 2 ways

Probability = 2/10 = 1/5

৯৬০.

By selling a pen for Tk 15, a man losses one-sixteenth of what it costs him. What is the cost price of the pen?

ক
ক) 16 Tk
খ
খ) 18 Tk
গ
গ) 21 Tk
ঘ
ঘ) 20 Tk

ব্যাখ্যা

Question: By selling a pen for Tk 15, a man losses one-sixteenth of what it costs him. What is the cost price of the pen?

Solution:
Let the cost price of the pen be Tk x

Then,
x - 15 = x/16
⇒ x - x/16 = 15
⇒ 15x/16 = 15
⇒ x/16 = 1
⇒ x = 16

৯৬১.

If ⁿC₇ = ⁿC₅ , then find the value of n.

ক
2
খ
12
গ
35
ঘ
50

ব্যাখ্যা

Question: If ⁿC₇ = ⁿC₅ , then find the value of n.

Solution:
We know,
If ⁿC_r = ⁿC_s , then, n = r + s

Now, ⁿC₇ = ⁿC₅
∴ n = 7 + 5
= 12

৯৬২.

A coin is thrown 3 times. What is the probability that atleast one head is obtained?

ক
1
খ
1/2
গ
3/8
ঘ
7/8
ঙ
1/8

ব্যাখ্যা

Question: A coin is thrown 3 times. What is the probability that atleast one head is obtained?

Solution:
Sample space = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Total number of ways = 2³ = 8.
Fav. Cases = 7

∴ Probability that atleast one head is obtained = 7/8

৯৬৩.

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

ক
ক) 1/15
খ
খ) 1/221
গ
গ) 25/57
ঘ
ঘ) 35/256
ঙ
ঙ) None of the above

ব্যাখ্যা

Let S be the sample space
Then,
n(S) = ⁵²C₂
= 52 × 51/2 × 1
= 1326
Let E = event of getting 2 kings out of 4.
n(E) = 4C2
=> 4 × 3/2 × 1
P(E) = n(E)/n(S)
= 6/1326
= 1/221

৯৬৪.

In how many different ways can be letters of the word 'CYCLE' be arranged?

ক
30 ways
খ
60 ways
গ
90 ways
ঘ
120 ways

ব্যাখ্যা

Question: In how many different ways can be letters of the word 'CYCLE' be arranged?

Solution:
CYCLE whereas total 5 letters and C comes two times.

So, arrangements are = 5!/2!
= 60 ways

৯৬৫.

There are 12 white and 6 yellow balls in a bag. Probability of drawing a white ball is -

ক
3/2
খ
2/3
গ
1/3
ঘ
4/3

ব্যাখ্যা

Total balls = 12 + 6 = 18
White balls = 12
Probability of drawing a white ball is 12/18 = 2/3