Question 1

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

Accepted Answer

P(odd) = P (even) = 1 2 1(because there are 50 odd and 50 even numbers) Sum or the three numbers can be odd only under the following 4 scenarios: Odd + Odd + Odd = 1 2 * 1 2 * 1 2 = 1 8 Odd + Even + Even = 1 2 * 1 2 * 1 2 = 1 8 Even + Odd + Even = 1 2 * 1 2 * 1 2 = 1 8 Even + Even + Odd = 1 2 * 1 2 * 1 2 = 1 8 Other combinations of odd and even will give even numbers. Adding up the 4 scenarios above: = 1 8 + 1 8 + 1 8 + 1 8 = 4 8 = 1 2

Question 2

A number is selected at random from the numbers 1 to 30. What is the probability that it is divisible by either 3 or 7 ?

Accepted Answer

1/30

Question 3

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

Accepted Answer

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 = 1 2 2 × 1 2 9 = 1 2 11

Question 4

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.

Accepted Answer

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 E ¯ = 52 - 16 = 36 Therefore, P E = 36 52 = 9 13

Question 5

The probability of occurance of two two events A and B are 1/4 and 1/2 respectively. The probability of their simultaneous occurrance is 7/50. Find the probability that neither A nor B occurs.

Accepted Answer

P ( neither A nor B) = P A a n d B = P A ∩ B = P A ∪ B = 1 - P A ∪ B = 1 - 61 100 = 39 100

Question 6

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

Accepted Answer

Let S be the sample space Then, n(S) = 52 C 2 = 1326. Let E = event of getting 2 kings out of 4. n(E) = 52 * 51 2 * 1 = 6. 4 C 2 => 4 * 3 2 * 1 P ( E ) = n ( E ) n ( S ) = 6 1326 = 1 221

Question 7

A card is drawn from a pack of 52 cards. The card is drawn at random. What is the probability that it is neither a spade nor a Jack?

Accepted Answer

There are 13 spade and 3 more jack Probability of getting spade or a jack: =13+352=1652=413 So probability of getting neither spade nor a jack: =1−413 = 9/13

Question 8

Three unbiased coins are tossed. What is the probability of getting at most two heads ?

Accepted Answer

Let S be the sample space. Here n(S)= 2 3 = 8 Let E be the event of getting atmost two heads. Then, n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)} Required probability = n(E)/n(S) = 6/8 = 3/4.

Question 9

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 15 ?

Accepted Answer

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20 Let E = event of getting a multiple of 4 or 15 =multiples od 4 are {4, 8, 12, 16, 20} And multiples of 15 means multiples of 3 and 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}. = the common multiple is only (15). => E = n(E)= 6 Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

Question 10

The Manager of a company accepts only one employees leave request for a particular day. If five employees namely Roshan, Mahesh, Sripad, Laxmipriya and Shreyan applied for the leave on the occasion of Diwali. What is the probability that Laxmi priya…

Accepted Answer

Number of applicants = 5 On a day, only 1 leave is approved. Now favourable events = 1 of 5 applicants is approved Probability that Laxmi priya's leave is granted = 1/5.

Question 11

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.

Accepted Answer

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 E 3 = E 1 ∩ E 2 = { 6 , 12 , 18 } => n(E3) = 3 ∴ E = E 1 ∪ E 2 = E 1 + E 2 - E 3 => n(E) = 9 + 6 - 3 =12 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 } ∴ P ( E ) = n ( E ) n ( S ) = 12 18 = 2 3

Question 12

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?

Accepted Answer

n(S) = C 2 52 = 1326 Let A = event of getting both red cards and B = event of getting both queens then A ∩ B = event of getting two red queens n(A) = C 2 26 = 325, n(B) = C 2 4 = 6 n ( A ∩ B ) = C 2 2 = 1 ∴ P A = 325 1326 , P B = 6 1326 P A ∩ B = 1 1326 ∴ P ( both red or both queens) = P A ∪ B = P A + P B - P A ∩ B = 325 1326 + 1 221 - 1 1326 = 55 221

Question 13

A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

Accepted Answer

Let number of balls = (6 + 8) = 14. Number of white balls = 8. P (drawing a white ball) = 8 /14=4/7.

Question 14

If p:q are the odds in favour of an event,then the probability of that event is:

Accepted Answer

Total number of cases=p+q Favourable cases=p Probability of that event is=p/(p+q)

Question 15

A three-digit number is formed with the digits 1, 2, 3, 4, 5 at random. What is probability that number formed is divisible by 5 ?

Accepted Answer

1/5

Question 16

The probabilities that drivers A, B and C will drive home safely after consuming liquor are 2 / 5, 3 / 7 and 3 / 4, respectively. What is the probability that they will drive home safely after consuming liquor ?

Accepted Answer

Let A be the event of driver A drive safely after consuming liquor. Let B be the event of driver B drive safely after consuming liquor. Let C be the event of driver C drive safely after consuming liquor. P(A)=2/5,P(B)=3/7,P(C)=3/4 The events A, B and C are independent . Therefore, P A ∩ B ∩ C = P ( A ) P ( B ) P ( C ) = 2 5 * 3 7 * 3 4 = 9 70 Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9/70

Question 17

The probability that a card drawn from a pack of 52 cards will be a diamond or a king, is

Accepted Answer

4/13

Question 18

Consider 5 independent Bernoulli’s trials each with probability of success p. If the probability of at least one failure is greater than or equal to 31/32 , then p lies in the interval ?

Accepted Answer

Probability of atleast one failure = 1 - no failure > 31/32 = 1 - P 5 > 31/32 = P 5 < 1/32 = p < 1/2 Also p > 0 Hence p lies in [0,1/2].

Question 19

There are 10 Letters and 10 correspondingly 10 different Address. If the letter are put into envelope randomly, then find the Probability that Exactly 9 letters will at the Correct Address ?

Accepted Answer

As we know we have 10 letter and 10 different address and one more information given that exactly 9 letter will at the correct address....so the remaining one letter automatically reach to their correct address P(E) = favorable outcomes /total outcomes Here favorable outcomes are '0'. So probability is '0'.

Question 20

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

Accepted Answer

Here, n(S) = 52. Let E = event of getting a queen of club or a king of heart. Then, n(E) = 2. P(E) =n(E)/n(S)=2/52=1/26.

Probability Questions