8
Q:

# A coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely. Find the probability that the die shows a number greater than 4, if it is known that the first throw of the coin results in a tail

 A) 1/3 B) 2/3 C) 2/5 D) 4/15

Explanation:

Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }

Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,

A = { T5, T6 }

B = { T1, T2, T3, T4, T5, T6 }

Therefore, Required probability = $P\left(\frac{A}{B}\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{2}{8}}{6}{8}}=\frac{1}{3}$

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) =

6 119
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 = $\left(\frac{\left(\mathbf{11}\mathbf{!}\mathbf{×}\mathbf{3}\mathbf{!}\right)}{\mathbf{13}\mathbf{!}}\right)$ = $\frac{39916800×6}{6227020800}$ = $\frac{\mathbf{24}}{\mathbf{625}}$

9 430
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.

21 580
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

11 550
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

8 559
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.

13 939
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

Explanation:

Probability of not hitting the ball = 1- 1/4 =
Then, the probability that the batsman will hit the ball =

10 821
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 the card drawn is red.

 A) 2/3 B) 1/2 C) 1/52 D) 13/51

Explanation:

We know that:

When one card is drawn from a pack of 52 cards

The numbers of possible outcomes n(s) = 52

We know that there are 26 red cards in the pack of 52 cards
⇒ The numbers of favorable outcomes n(E) = 26

Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)

∴ required probability = 26/52 = 1/2.