14
Q:

# A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

 A) 23/42 B) 19/42 C) 7/32 D) 16/39

Explanation:

A red ball can be drawn in two mutually exclusive ways

(i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore

P(E1) = 1/2  and  P(E2) = 1/2

Now, $\inline P(\frac{A}{E1})$ = Probability of drawing a red ball when the first bag has been selected = 4/7

$\inline P(\frac{A}{E2})$  = Probability of drawing a red ball when the second bag has been selected = 2/6

Using the law of total probability, we have

P(red ball) = P(A) = $\inline P(E1)P(\frac{A}{E1})+P(E2)P(\frac{A}{E2})$

= $\inline \frac{1}{2}\times \frac{4}{7}+\frac{1}{2}\times \frac{2}{6}=\frac{19}{42}$

Q:

A fair six-sided die is rolled twice. What is the probability of getting 4 on the first roll and not getting 6 on the second roll ?

 A) 1/36 B) 5/36 C) 1/12 D) 1/9

Explanation:

The two events mentioned are independent.

The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.

P(getting first 4) = 1/6

P(no second 6) = 5/6

Therefore P(getting first 4 and no second 6) = 1/6 x 5/6 = 5/36

1 46
Q:

Let K and L be events on the same sample space, with P (K) = 0.8 and P (B) = 0.6. Are these two events are disjoint ?

 A) TRUE B) FALSE

Explanation:

These two events cannot be disjoint because P(K) + P(L) > 1.

P(AꓴB) = P(A) + P(B) - P(AꓵB).

An event is disjoint if P(A ꓵ B) = 0. If K and L are disjoint P(K ꓴ L) = 0.8 + 0.6 = 1.4

And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

1 16
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

5 266
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

3 221
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.