ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Here S = {TTT, TTH,THT, HTT, THH, HTH, HHT, HHH}.
Let,
E = event of getting at least two heads
= {THH, HTH, HHT, HHH}.
∴ P(E) = n(E)/n(S)
= 4/8
= 1/2.

Total number of balls = (4 + 5 + 6)
= 15.
P(drawing a red ball or a green ball)
= P(red) + P(green)
= (4/15 + 5/15)
= 9/15
= 3/5.

Total number of balls = (6 + 4 + 2 + 3)
= 15.
Let,
E be the event of drawing 2 red balls.
Then,
n(E) = ⁶C₂
= (6 × 5)/(2 × 1)
= 15.
Also, n(S) = ¹⁵C₂
= (15 × 14)/(2 × 1)
= 105.
∴ P(E) = n(E)/n(S)
= 15/105
= 1/7.

Let,
S be the sample space and
E be the event of select.
Then,
n(S) = number of ways of selecting 3 students out of 25
= ²⁵C₃
= (25 × 24 × 23)/(3 × 2 × 1)
= 2300.
And,
n(E) = (¹⁵C₂ × ¹⁰C₁)
= {(15 × 14)/(2 × 1)} × 10
= 1050.
∴ P(E) = n(E)/n(S)
= 1050/2300
= 21/46.

Number of red balls = 4
Number of yellow ball = 5
Number of pink ball = 6
Total number of balls = 4 + 5 + 6 = 15
Total possible outcomes = selection of 2 balls out of 15 balls = ¹⁵C₂
= 15!/2!(15 - 2)!
= 15!/(2! × 13!)
= (15 × 14)/(1 × 2)
= 105.
Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls.
¹⁰C₂
= 10!/2!(10 - 2)!
= 10!/2!8!
= (10 × 9)/(1 × 2)
= 45.
∴ Required Probability = 45/105
= 3/7

Total results = 2x + 3x + 4x = 9x,
Favourable result is (red ball) = 2x
∴ Probability = 2x/9x
= 2/9
Answer: 2/9

Here,
Total results S = {1, 2, 3, 4, .....19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}
∴ P(E) = n(E)/n(S)
= 9/20
= 4.5/10
= 0.45
Answer: 0.45

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।

Let S be the sample space.
Total number of students in the class=12 boys + 16 girls = 28
Then, n(S) = 28
Let E be the event of calling one of them by enrollment number.
Given that, the number of girls = 16.
Then, n(E) = 16.
The probability that the one called is a girl = n(S)/n(E) = 16/28 = 4/7.

Total number of drink bottles = 6 + 3 + 4 = 13.
Let S be the sample space
Then, n(S) = number of ways of taking 3 drink bottles out of 13.
Therefore, n(S) = ¹³C₃
= (13 x 12 x 11)/(1 x 2 x 3)
= 66 x 13
= 858.
Let E be the event of taking 3 bottles of the same variety
Then, E = event of taking (3 bottles out of 6) or (3 bottles out of 3) or (3 bottles out of 4)
n(E) = ⁶C₃ + ³C₃ + ⁴C₃
= (6 x 5 x 4 )/ (1 x 2 x 3) + 1 + (4 x 3 x 2) / (1 x 2 x 3)
= 20 + 1 + 4
= 25
The probability of taking 3 bottles of the same variety = n(E)/n(S)
= 25/858.
Then, the probability of taking 3 bottles are not of the same variety = 1 - 25/858
= 833/858.

Total number of students = 20.
Let S be the sample space.
Then, n(S) = number of ways of three scored first mark
n(S) = ²⁰C₃
= 20 x 19 x 18 / 2 x 3
= 20 x 19 x 3

Let,
E be the event of 1 girl and 2 boys.
Therefore, n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12.
n(E) = ⁸C₁ x ¹²C₂
= 8 x 12 x 11 / 1 x 2
= 8 x 6 x 11.

Now, the required probability = n(E)/n(S)
= (8 x 6 x 11)/(20 x 19 x 3)
= 44/95.

Let, A be the event of the group A pass
Let, B be the event of the group B pass

Then,
A'= Event of the group A's fail and B'= event of the group B's fail.
Therefore, p(A) = 2/7 and p(B) = 2/5,
P(A') = 1 - P(A) = 1- 2/7 = 5/7 and P(B') = 1- P(B) = 1- 2/5 = 3/5

Required probability = P[( A And B') Or (B And A')]
= P[( A And B') Or (B And A')]
= P[( A And B') + (B And A')]
= P[( A And B')] + P[(B And A')]
= p(A) x P(B') + P(A') x P(B)
= (2/7 x 3/5) + (2/5 x 5/7)
= (6/35 + 10/35)
= 16/35

The given word contains 8 different letters.
When the vowels AUE are taken together, we may treat them as 1 letter.
Then,
The letters to be arranged are DGHTR (AUE)
The letters can be arranged in ⁶P₆ = 6!
= 720 ways.
The vowels AUE may be arranged in 3! = 6 ways.
Required number of ways = (720 × 6)
= 4320 ways.

Required number of ways = ⁶C₁× ⁸C₃ + ⁶C₂× ⁸C₂ + ⁶C₃× ⁸C₁+ ⁶C₄× ⁸C₀
= {6 × (8 × 7 × 6)/3!} + {(6 × 5)/(2 × 1) × (8 × 7)/(2 × 1)} + {(6 × 5 × 4)/3! × 8} + {(6 × 5)/(2 × 1) × 1}
= (336 + 420 + 160 + 15)
= 931.

The given word contains 7 different letters.
Keeping the vowels (AUIO) together, we take them as 1 letter.
Then,
we have to arrange the letters CTN(AUIO).
Now, 4 letters can be arranged in 4! = 24 ways.
The vowels (AUIO) can be arranged themselves in 4! = 24 ways.
∴ Required number of ways = (24 × 24)
= 576.

যে কোন করমর্দন অথবা কোলাকুলির অংকে শুধু কত জন করমর্দন (Handshake), বা কোলাকুলি করল তা দেয়া থাকবে।
এক্ষেত্রে মনে রাখতে হবে যে প্রত্যেক বার করমর্দন বা কোলাকুলি করার সময় মোট ২ জন লোকের প্রয়োজন।
তাই এক্ষেত্রে সূত্রটি হবে ⁿC₂ = ^{মোট লোক}C_{২ জন সব সময়}
¹⁰C₂ = 10!/2!(10 - 2)!
= 10!/2!8!
= (10 × 9)/2
= 5 × 9
= 45.

Here, the order of arrangement of digits does matter.
ⁿP_r = n!/(n-r)!
ⁿP_r = 4!/(4-3)!
⁴P₃ = 4!/1!
⁴P₃ = 4!
⁴P₃ = 24

ⁿP_r = n!/(n-r)!
⁶P₃ = 6!/(6-3)!
⁶P₃ = 6!/3!
⁶P₃ = 120

Required no. of numbers = 5 ×⁵P₄
= 5 × 5!
= 5 ×120
= 600

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= ⁷C₃× ⁴C₂
= {(7 × 6 × 5)/(3 × 2 × 1)} × {(4 × 3)/(2 × 1)}
= 210.

Clearly,
n(S) = 6 × 6
= 36.
Let,
E be the event that the sum of the numbers on the two faces is divisible by 4 or 6.
Then, E = {(1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (6, 2), (6, 6)}
∴ n(E) = n(E)/n(S)
= 14/36
= 7/18.