2
Q:

# A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

 A) 1/6 B) 1/3 C) 1/2 D) 1/4

Explanation:

P(odd) = P (even) =$\frac{1}{2}$ 1(because there are 50 odd and 50 even numbers)

Sum or the three numbers can be odd only under the following 4 scenarios:

Odd + Odd + Odd = $\frac{1}{2}*\frac{1}{2}*\frac{1}{2}$$\frac{1}{8}$

Odd + Even + Even = $\frac{1}{2}*\frac{1}{2}*\frac{1}{2}$=$\frac{1}{8}$

Even + Odd + Even = $\frac{1}{2}*\frac{1}{2}*\frac{1}{2}$=$\frac{1}{8}$

Even + Even + Odd = $\frac{1}{2}*\frac{1}{2}*\frac{1}{2}$ = $\frac{1}{8}$

Other combinations of odd and even will give even numbers.

Adding up the 4 scenarios above:

$\frac{1}{8}$$\frac{1}{8}$+$\frac{1}{8}$$\frac{1}{8}$ = $\frac{4}{8}$ = $\frac{1}{2}$

Q:

In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin?

 A) 4/7 B) 2/3 C) 1/2 D) 5/6

Explanation:

Total coins 30

In that,

1 rupee coins 20

50 paise coins 10

Probability of total 1 rupee coins =  20C11

Probability that 11 coins are picked = 30C11

Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.

0 3
Q:

In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?

 A) 15 B) 18 C) 20 D) 24

Explanation:

We know that, Total probability = 1

Given probability of black stones = 1/4

=> Probability of blue and white stones = 1 - 1/4 = 3/4

But, given blue + white stones =  9 + 6 = 15

Hence,

3/4 ----- 15

1   -----  ?

=> 15 x 4/3 = 20.

Hence, total number of stones in the box = 20.

5 305
Q:

What is the probability of an impossible event?

 A) 0 B) -1 C) 0.1 D) 1

Explanation:

The probability of an impossible event is 0.

The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.

The probability of a certain event is 1.

8 730
Q:

In a box, there are four marbles of white color and five marbles of black color. Two marbles are chosen randomly. What is the probability that both are of the same color?

 A) 2/9 B) 5/9 C) 4/9 D) 0

Explanation:

Number of white marbles = 4

Number of Black marbles = 5

Total number of marbles = 9

Number of ways, two marbles picked randomly = 9C2

Now, the required probability of picked marbles are to be of same color = 4C2/9C2  +  5C2/9C2

= 1/6 + 5/18

= 4/9.

7 1039
Q:

A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red?

 A) 2/3 B) 1/8 C) 3/8 D) 3/4

Explanation:

Given number of balls = 3 + 5 + 7 = 15

One ball is drawn randomly = 15C1

probability that it is either pink or red =

13 968
Q:

Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?

 A) 1/4 B) 1/6 C) 1/8 D) 4

Explanation:

Required probability is given by P(E) =

14 1512
Q:

14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.

 A) 11/379 B) 21/628 C) 24/625 D) 26/247

Explanation:

Total no of ways = (14 – 1)! = 13!

Number of favorable ways = (12 – 1)! = 11!

So, required probability = $\left(\frac{\left(\mathbf{11}\mathbf{!}\mathbf{×}\mathbf{3}\mathbf{!}\right)}{\mathbf{13}\mathbf{!}}\right)$ = $\frac{39916800×6}{6227020800}$ = $\frac{\mathbf{24}}{\mathbf{625}}$

15 1474
Q:

Two dice are rolled simultaneously. Find the probability of getting the sum of numbers on the on the two faces divisible by 3 or 4?

 A) 3/7 B) 7/11 C) 5/9 D) 6/13

Explanation:

Here n(S) = 6 x 6 = 36

E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}

=> n(E)=20

Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.