3
Q:

# Tickets are numbered from 1 to 18 are mixed up together and then 9 tickets are drawn at random. Find the probability that the ticket has a number, which is a multiple of 2 or 3.

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

Explanation:

S = { 1, 2, 3, 4, .....18 }

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

$\inline E3=(E1\cap E2) = \left \{ 6,12,18 \right \}$

=> n(E3) = 3

$\inline \therefore E = E1 \cup E2=E1+E2-E3$

=> n(E) = 9 + 6 - 3 =12

where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

$\inline \therefore P(E)= \frac{n(E)}{n(S)}= \frac{12}{18}=\frac{2}{3}$

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.

13 117
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

8 133
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

7 229
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.

7 598
Q:

The probability that a bowler bowled a ball from a point will hit by the batsman is ¼. Three such balls are bowled simultaneously towards the batsman from that very point. What is the probability that the batsman will hit the ball ?

 A) 37/64 B) 27/56 C) 11/13 D) 9/8