11
Q:

# If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.

 A) 5/12 B) 7/12 C) 3/14 D) 1/12

Explanation:

Total number of elementary events = $10{C}_{5}$

Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is $7{C}_{5}$.

So,required probability =$7{C}_{5}$$10{C}_{5}$ = 1/12.

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.

2 115
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.

1 396
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 =

7 379
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) =

12 684
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 = $\left(\frac{\left(\mathbf{11}\mathbf{!}\mathbf{×}\mathbf{3}\mathbf{!}\right)}{\mathbf{13}\mathbf{!}}\right)$ = $\frac{39916800×6}{6227020800}$ = $\frac{\mathbf{24}}{\mathbf{625}}$

12 881
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.

26 1033
Q:

A person starting with 64 rupees and making 6 bets, wins three times and loses three times, the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is:

 A) A gain of Rs. 27 B) A loss of Rs. 37 C) A loss of Rs. 27 D) A gain of Rs. 37

Explanation:

As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order).

The overall resultant will remain same.

So final amount with the person will be (in all cases):

64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27

Hence the final result is:

64 − 27 37

A loss of Rs.37

14 1151
Q:

A card is drawn from a pack of 52 cards. The probability of getting a queen of the club or a king of the heart is?

 A) 1/26 B) 1/13 C) 2/13 D) 1/52

Explanation:

Here in this pack of cards, n(S) = 52

Let E = event of getting a queen of the club or a king of the heart

Then, n(E) = 2

P(E) = n(E)/n(S) = 2/52 = 1/26