ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

সিলেবাস

"Exam - 54

Daily Quiz
Math:
Topic: Probability, Permutation and Combination"

প্রশ্ন দেখুন অনুশীলন

ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ] · তারিখ অনির্ধারিত · ১৪ প্রশ্ন

১.

Tickets numbered 1 to 24 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 4?

ক
11/24
খ
1/2
গ
5/12
ঘ
3/8

সঠিক উত্তর: খ

1/2

উত্তর

সঠিক উত্তর: খ

1/2

খ

ব্যাখ্যা

Question: Tickets numbered 1 to 24 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 4?

Solution:
Here, S = {1, 2, 3, 4, ...., 23, 24}
Let E = event of getting a multiple of 3 or 4
= {3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21, 24}

∴P(E) = n(E)/n(S)
= 12/24
= 1/2

২.

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

ক
60
খ
320
গ
420
ঘ
720

সঠিক উত্তর: গ

420

উত্তর

সঠিক উত্তর: গ

420

গ

ব্যাখ্যা

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

Solution:
The word 'SUCCESS' consists of 7 letters.
Here,
Number of letters, S = 3
Number of letters, C = 2

∴ Number of arrangements = 7!/(3! × 2!)
= 5040/12
= 420

৩.

In how many ways can 8 Bangladeshi, 4 American, and 4 Japanese be seated in a row so that all people of the same nationality sit together?

ক
8! 4! 4!
খ
8!/3! 4!
গ
8! 4! 3!
ঘ
3! 8! 4! 4!

সঠিক উত্তর: ঘ

3! 8! 4! 4!

উত্তর

সঠিক উত্তর: ঘ

3! 8! 4! 4!

ঘ

ব্যাখ্যা

Question: In how many ways can 8 Bangladeshi, 4 American, and 4 Japanese be seated in a row so that all people of the same nationality sit together?

Solution:
Taking all people of the same nationality as one person, then we will have only three people.
These three people can be arranged themselves in = 3! Ways
8 Bangladeshi can be arranged themselves in = 8! Ways
4 Americans can be arranged themselves in = 4! Ways
4 Japanese can be arranged themselves in = 4! Ways

Hence, the required number of ways = 3! 8! 4! 4! Ways

৪.

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

সঠিক উত্তর: ক

1260

উত্তর

সঠিক উত্তর: ক

1260

ক

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

৫.

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

সঠিক উত্তর: গ

1/16

উত্তর

সঠিক উত্তর: গ

1/16

গ

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

৬.

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

সঠিক উত্তর: গ

100

উত্তর

সঠিক উত্তর: গ

100

গ

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

৭.

A company offers a bonus to employees who complete certain training modules. The probability that Emma will complete the "Leadership Skills" module is 0.8, and the probability that John will complete the "Time Management" module is 0.5. What is the probability that both Emma and John will complete their respective modules?

ক
0.4
খ
0.75
গ
0.3
ঘ
0.6

সঠিক উত্তর: ক

0.4

উত্তর

সঠিক উত্তর: ক

0.4

ক

ব্যাখ্যা

Question: A company offers a bonus to employees who complete certain training modules. The probability that Emma will complete the "Leadership Skills" module is 0.8, and the probability that John will complete the "Time Management" module is 0.5. What is the probability that both Emma and John will complete their respective modules?

Solution:
Let,
Probability that Emma completes the "Leadership Skills" module P(E) = 0.8
P(J) = Probability that John completes the "Time Management" module P(J) = 0.5

Since the events are independent, the probability that both Emma and John will complete their respective modules,
P (E ∩ J) = P(E) × P (J)
= 0.8 × 0.5
= 0.4

৮.

In a football championship, there are 15 matches. If each team plays one match with every other team, the number of teams is-

ক
5
খ
6
গ
7
ঘ
8

সঠিক উত্তর: খ

উত্তর

সঠিক উত্তর: খ

খ

ব্যাখ্যা

Question: In a football championship, there are 15 matches. If each team plays one match with every other team, the number of teams is-

Solution:
Let n be the number of teams.
ⁿC₂ = 15
⇒ n!/{2! × (n - 2)!} = 15
⇒ {n(n - 1)(n - 2)!}/{2! × (n - 2)!} = 15
⇒ n(n - 1)/2 = 15
⇒ n(n - 1) = 30
⇒ n2 - n - 30 = 0
⇒ n - 6n + 5n - 30 = 0
⇒ n(n - 6) + 5(n - 6) = 0
⇒ (n - 6)(n + 5)
∴ n = 6 [Negative value is not acceptable]

৯.

What is the probability of getting a sum of 8 when two dice are thrown?

ক
1/6
খ
5/36
গ
1/9
ঘ
7/36

সঠিক উত্তর: খ

5/36

উত্তর

সঠিক উত্তর: খ

5/36

খ

ব্যাখ্যা

Question: What is the probability of getting a sum of 8 when two dice are thrown?

Solution:
Total number of ways = 6 × 6 = 36 ways.
Favorable cases = (2, 6), (6, 2), (3, 5), (5, 3), (4, 4) = 5 ways.

∴ Probability = 5/36

১০.

Three gentlemen and three ladies are candidates for two vacancies. A voter has to vote for two candidates. In how many ways can one cast his vote?

ক
10
খ
12
গ
15
ঘ
18

সঠিক উত্তর: গ

উত্তর

সঠিক উত্তর: গ

গ

ব্যাখ্যা

Question: Three gentlemen and three ladies are candidates for two vacancies. A voter has to vote for two candidates. In how many ways can one cast his vote?

Solution:
ways can one cast his vote =⁶C₂
= 15

১১.

If the probability of rain on any given day in City Cumilla is 50%, what is the probability that it rains on exactly 2 days in a 5-day period?

ক
7/20
খ
5/16
গ
2/5
ঘ
4/9

সঠিক উত্তর: খ

5/16

উত্তর

সঠিক উত্তর: খ

5/16

খ

ব্যাখ্যা

Question: If the probability of rain on any given day in City Cumilla is 50%, what is the probability that it rains on exactly 2 days in a 5-day period?

Solution:
If the probability of rain on any given day in City Cumilla is 50%
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 5 = ⁵C₂

∴the probability that it rains on exactly 2 days in a 5-day period is = ⁵C₂ × (1/2) × (1/2) × (1/2) × (1/2) × (1/2)
= 10 × (1/32)
= 5/16

১২.

In how many ways can 5 examination papers be arranged so that the best and the worst papers never come together?

ক
58 ways
খ
66 ways
গ
72 ways
ঘ
76 ways

সঠিক উত্তর: গ

72 ways

উত্তর

সঠিক উত্তর: গ

72 ways

গ

ব্যাখ্যা

Question: In how many ways can 5 examination papers be arranged so that the best and the worst papers never come together?

Solution:
Total ways = 5!
= 120 ways

if two papers come together, we can consider them one.
ways that they will come together = 4! × 2!
= 24 × 2
= 48 ways

∴ ways the best and the worst papers never come together = 120 - 48 ways
= 72 ways

১৩.

A box contains 100 pens. Out of which six are defective. Four pen is out from the box. Find the probability that the pen is not defective.

ক
1/10
খ
9/10
গ
8/49
ঘ
47/50

সঠিক উত্তর: ঘ

47/50

উত্তর

সঠিক উত্তর: ঘ

47/50

ঘ

ব্যাখ্যা

Question: A box contains 100 pens. Out of which six are defective. Four pen is out from the box. Find the probability that the pen is not defective.

Solution:
Total pen = 100
Good pen = 100 - 6 = 94

The probability of pen is not defective = 94/100
= 47/50

১৪.

Find the probability that a leap year has 52 Sundays.

ক
3/8
খ
5/7
গ
3/13
ঘ
5/14

সঠিক উত্তর: খ

5/7

উত্তর

সঠিক উত্তর: খ

5/7

খ

ব্যাখ্যা

Question: Find the probability that a leap year has 52 Sundays.

Solution:
A leap year can have 52 Sundays or 53 Sundays.
In a leap year, there are 366 days out of which there are 52 complete weeks & remaining 2 days.

Now, these two days can be (Sat, Sun) (Sun, Mon) (Mon, Tue) (Tue, Wed) (Wed, Thur) (Thur, Friday) (Friday, Sat).
So there are total 7 cases out of which (Sat, Sun) (Sun, Mon) are two favorable cases.
So, P(53 Sundays) = 2/7

Now,
P(52 Sundays) + P(53 Sundays) = 1
So, P(52 Sundays) = 1 - P(53 Sundays) = 1 - (2/7) = 5/7