ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

In two throws of a dice, n(S) = (6 x 6) = 36
Let E = event of getting a sum
         = {(3, 6), (4, 5), (5, 4), (6, 3)}
∴P(E) = n(E)/n(S)
           =4/36
           =1/9

Total number of balls
= (8 + 7 + 6)
= 21
Let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue
∴n(E)=7
∴P(E) = n(E)/n(S)
          = 7/21
          = 1/3

Getting at most Two heads means 0 to 2 but not more than 2
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
∴P(E) = n(E)/n(S) = 7/8

Let the probability of winning the race is denoted by P(person)
P(P)=1/4,P(Q)=1/5,P(R)=1/6,P(S)=1/7
All the events are mutually exclusive (since if one of them wins then other would lose as pointed out by rahul) hence,
Required probability:
= P(P) + P(Q) + P(R) + P(S)
=1/4 + 1/5 + 1/6 + 1/7 = 319/420

If 2 marbles are picked at random from 6 red marbles,
then the probability
= ⁶C₂
= 15

If 2 marbles are picked at random from (6 + 5 + 2) or 13 marbles,
then the probability
= ¹³C₂
= 78

P(Both are red)
= 15/78
= 5/26

One digit positive numbers = 5
Two digit positive numbers = 25
Three digit positive numbers = 100
4 digit positive numbers = 300
5 digit positive numbers = 600
Six digit positive numbers = 600
Total positive numbers,
= 5 + 25 + 100 + 300 + 600 + 600
= 1630
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Alternative way:

Let n be the number of persons in the party
Number of hands shake = 105
Total number of hands shake is given by ⁿC₂
Now,
According to the question,
ⁿC₂ = 105
or, n!/(2!×(n−2)!) = 105
or, n × (n−1)/2 = 105
or, n² − n = 210
or, n² − n − 210 = 0
or, n = 15,−14

But, we cannot take negative value of n
So, n = 15

The number of triangle can be formed by 10 points = ¹⁰C₃
Similarly, the number of triangle can be formed by 4 points when no one is collinear = ⁴C₃
In the question, given 4 points are collinear, Thus, required number of triangle can be formed,
= ¹⁰C₃ - ⁴C₃
= 120 - 4
= 116

To committee can be formed in the following ways,
(1 lady + 4 gents) or (2 ladies + 3 gents) or (3 ladies + 2 gents) or (4 ladies + 1 gents) or (5 ladies + 0 gents).
Total number of possible arrangements,
= (⁴C₁ × ⁶C₄) + (⁴C₂ × ⁶C₃) + (⁴C₃ × ⁶C₂) + (⁴C₄ × ⁶C₁)
= 60 + 120 + 60 + 6
= 246



[ From 6 men and 4 ladies, a committee of 5 is to be formed. In how many ways can this be done, if there is no restriction about its formation?
Solution:
From 6 men and 4 ladies, a committee of 5 is to be formed.
Total number of possible arrangements, if there is no restriction about its formation
= ^{(6 + 4)}C₅
= ¹⁰C₅
= 252 ]

ⁿP_r = ⁿC_r × r!
⁶P_r = 15 × r!
360 = 15 × r!
r! = 360/15
 = 24
r! = 4 × 3 × 2 × 1
⇒ r! = 4!
Therefore, r = 4
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Alternative way:
⁶P_r/⁶C_r = 360/15
or, 6!/(6 - r)! ÷ 6!/{r!(6 - r)!} = 24
or, r! = 4!
∴ r = 4

3 prize among 5 students can be distributed in ⁵C₃ ways = 10 ways.

There are 8 students and the maximum capacity of the cars together is 9.
We may divide the 8 students as follows:

Case I:
5 students in the first car and 3 in the second.
Hence, 8 students are divided into groups of 5 and 3
in ⁸C₃ = 56 ways.

Case II:
4 students in the first car and 4 in the second.
So, 8 students are divided into two groups of 4 and 4
in ⁸C₄ = 70 ways.

Therefore, the total number of ways in which 8 students can travel is:
56 + 70 = 126

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 2 blue and 1 yellow marble.
Then, n(E) = ⁴C₂ × ³C₁
   =(4 × 3)/(2 × 1) × 3
   = 18
Also, n(S) = ¹⁵C₃
    = 15 × 14 × 13/(3 × 2 × 1)
    = 455
∴P(E) = n(E)/n(S) = 18/455

Here, n(S) = 52
There are 13 cards of diamond (including one king) and there are three more kings.
Let E = event of getting a diamond or a king
Then, n(E) = (13 + 3) = 16
∴P(E) = n(E)/n(S)
          =16/52
          = 4/13

P(E) = 30/100 = 3/10;
P(B) = 20/100 = 1/5
and P(E∩B) = 10/100 =1/10
P (E or B)
= P(E∪B)
= P(E) + P(B) - P(E∪B)
= (3/10+1/5−1/10)
= 4/10
= 2/5

APPLE = 5 letters.
But two letters PP is of same kind.
Thus, required permutations,
= 5!/2!
= 120/2
= 60

Taking all person of same nationality as one person, then we will have only three people.
These three person can be arranged themselves in 3! Ways.
8 Bangladeshi can be arranged themselves in 8! Way.
4 American can be arranged themselves in 4! Ways.
4 Englishman can be arranged themselves in 4! Ways.
Hence, required number of ways = 3! 8! 4! 4! Ways.
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Alternative way:

We are selecting three different posts here, so order matters.
Thus, total ways of selecting a President, VP and Water-boy from a group of 10 people would be ¹⁰P₃
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Alternative way:

There are 9 non-zero digits to arrange themselves at 4 different position.
Each letter can be arrange at different position in 9 different ways.
So, required number of ways,
= 9 × 9 × 9 × 9
= 9^{4
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