12
Q:

# A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

 A) 136/345 B) 17/87 C) 316/435 D) 158/435

Explanation:

$\inline n(S)=^{30}\textrm{C}_{2}$

Let A be the event of getting two oranges and

B be the event of getting two non-defective fruits.

and $\inline (A\cap B)$ be the event of getting two non-defective oranges

$\inline \therefore$ $\inline P(A)=\frac{^{20}\textrm{C}_{2}}{^{30}\textrm{C}_{2}}, \; \; P(B)=\frac{^{22}\textrm{C}_{2}}{^{30}\textrm{C}_{2}} \; \; and\; \; P(A\cap B)= \frac{^{15}\textrm{C}_{2}}{^{30}\textrm{C}_{2}}$

$\inline \therefore P(A\cup B)= P(A)+P(B)-P(A\cap B)$

$\inline \frac{^{20}\textrm{C}_{2}}{^{30}\textrm{C}_{2}}+\frac{^{22}\textrm{C}_{2}}{^{30}\textrm{C}_{2}}-\frac{^{15}\textrm{C}_{2}}{^{30}\textrm{C}_{2}}=\frac{316}{435}$

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

4 85
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

3 87
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.

2 119
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 92
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