50
Q:

# A letter iws takenout at random from 'ASSISTANT'  and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

 A) 35/96 B) 19/90 C) 19/96 D) None of these

Explanation:

$\inline ASSISTANT\rightarrow AA\: I\: N\: SSS\: TT$

$\inline STATISTICS\rightarrow A\: II\: C\: SSS\: TTT$

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A =  $\inline \frac{^{2}\textrm{C}_{1}}{^{9}\textrm{C}_{1}}\times \frac{^{1}\textrm{C}_{1}}{^{10}\textrm{C}_{1}}$ = 1/45

Probability of choosing I = $\inline \frac{1}{^{9}\textrm{C}_{1}}\times \frac{^{2}\textrm{C}_{1}}{^{10}\textrm{C}_{1}}$ = 1/45

Probability of choosing S = $\inline \frac{^{3}\textrm{C}_{1}}{^{9}\textrm{C}_{1}}\times \frac{^{3}\textrm{C}_{1}}{^{10}\textrm{C}_{1}}$ = 1/10

Probability of choosing T = $\inline \frac{^{2}\textrm{C}_{1}}{^{9}\textrm{C}_{1}}\times \frac{^{3}\textrm{C}_{1}}{^{10}\textrm{C}_{1}}$ = 1/15

Hence, Required probability = $\inline \frac{1}{45}+\frac{1}{45}+\frac{1}{10}+\frac{1}{15}=\frac{19}{90}$

Q:

What is the probability that a leap year has 53 Saturdays and 52 Sundays ?

 A) 1/7 B) 2/7 C) 1/2 D) 3/2

Explanation:

A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Friday, Saturday}

Required Probability = 1/7

3 76
Q:

Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is  ?

 A) 16/19 B) 1 C) 3/2 D) 17/20

Explanation:

n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
P(E U P) = 10/20 + 8/20 - 1/20 = 17/20

2 79
Q:

The probability that a number selected at random from the first 100 natural numbers is a composite number is  ?

 A) 3/2 B) 2/3 C) 1/2 D) 34/7

Explanation:

The number of exhaustive events = 100 C₁ = 100.

We have 25 primes from 1 to 100.

Number of favourable cases are 75.

Required probability = 75/50 = 3/2.

2 109
Q:

In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples ?

 A) 6/7 B) 19/21 C) 7/31 D) 5/21

Explanation:

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ x ⁶C₁)

As remaining person can be any one among three couples left.

Required probability = (⁵C₂ x ⁶C₁)/¹⁰C₅
= (10 x 6)/252 = 5/21

1 80
Q:

If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is  ?

 A) 1 B) 1/2 C) 0 D) 3/5