5
Q:

Two cards are drawn at random from a well - shuffled pack of 52 cards. what is the probability that either both are red or both are queens?

 A) 17/112 B) 55/221 C) 55/121 D) 33/221

Explanation:

n(S) = ${}^{52}C_{2}$ = 1326

Let  A = event of getting both red cards

and B = event of getting both queens

then $\left(A\cap B\right)$ = event of getting two red queens

n(A) = ${}^{26}C_{2}$ = 325,   n(B) = ${}^{4}C_{2}$ = 6

$n\left(A\cap B\right)={}^{2}C_{2}=1$

$P\left(A\cap B\right)=\frac{1}{1326}$

$\therefore$P ( both red or both queens) = $P\left(A\cup B\right)$

$P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=$$\frac{325}{1326}+\frac{1}{221}-\frac{1}{1326}$=$\frac{55}{221}$

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

7 130
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.

21 373
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

10 287
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

7 394
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.

13 765
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 =

10 651
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.

7 806
Q:

A fair coin is tossed 11 times. What is the probability that only the first two tosses will yield heads ?

 A) (1/2)^11 B) (9)(1/2) C) (11C2)(1/2)^9 D) (1/2)

Explanation:

Probability of occurrence of an event,

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

⇒ Probability of getting head in one coin = ½,

⇒ Probability of not getting head in one coin = 1- ½ = ½,

Hence,

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads =${\left(\frac{1}{2}\right)}^{2}×{\left(\frac{1}{2}\right)}^{9}={\left(\frac{1}{2}\right)}^{11}$