2
Q:

# One card is drawn from a pack of 52 cards , each of the 52 cards being equally likely to be drawn. Find the probability that the card  drawn is neither a spade nor a king.

 A) 0 B) 9/13 C) 1/2 D) 4/13

Explanation:

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus   n(E) = 13 + 3 = 16

Hence $n\left(\overline{E}\right)=52-16=36$

Therefore, $P\left(\overline{)E}\right)=\frac{36}{52}=\frac{9}{13}$

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.

0 176
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 =

5 210
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) =

9 411
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 = $\left(\frac{\left(\mathbf{11}\mathbf{!}\mathbf{×}\mathbf{3}\mathbf{!}\right)}{\mathbf{13}\mathbf{!}}\right)$ = $\frac{39916800×6}{6227020800}$ = $\frac{\mathbf{24}}{\mathbf{625}}$

10 626
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.

23 768
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

12 857
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

8 694
Q:

A room contains 3 brown, 5 black and 4 white chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is white ?

 A) 14/33 B) 14/55 C) 12/55 D) 13/33

Explanation:

Total number of chairs = (3 + 5 + 4) = 12.

Let S be the sample space.

Then, n(s)= Number of ways of picking 2 chairs out of 12

12×11/2×66

Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.

=> 8C8×7/2×28

Therefore required probability = 28/66 = 14/33.