41
Q:

# A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

 A) 1/4 B) 1/2 C) 3/4 D) 7/12

Explanation:

Let A, B, C be the respective events of solving the problem and $\inline \overline{A},\overline{B},\overline{C}$ be the respective events of not solving the problem. Then A, B, C are independent events

$\inline \therefore \bar{A},\bar{B},\bar{C}$ are independent events

Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

$\inline P(\bar{A})=\frac{1}{2},P(\bar{B})=\frac{2}{3} \: and\: P(\bar{C})=\frac{3}{4}$

$\inline \therefore$ P( none  solves the problem) = P(not A) and (not B) and (not C)

= $\inline P(\bar{A}\cap \bar{B}\cap \bar{C})$

= $\inline P(\bar{A})P(\bar{B})P(\bar{C})$             $\inline (\because \bar{A},\bar{B}\: and\: \bar{C}\: are\: independent)$                                  =  $\inline \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}$

= $\inline \frac{1}{4}$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $\inline 1-\frac{1}{4}=\frac{3}{4}$

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