Question 1

A bag contains 2 red Roses, 4 yellow Roses and 6 pink Roses. Two roses are drawn at random. What is the probability that they are not of same color?

Accepted Answer

Possible outcomes = (RY, YP, PR) 2C1 4C1 + 4C1 6C1 + 6C1 2C1 Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2) = 8 + 24 + 12/66 = 44/66 = 2/3.

Question 2

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

Accepted Answer

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let E = event of getting at most two heads. Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}. P(E) =n(E)/n(S)=7/8.

Question 3

A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.

Accepted Answer

Total number of elementary events = 50 C 5 Given,third ticket =30 => first and second should come from tickets numbered 1 to 29 = 29 C 2 ways and remaining two in 20 C 2 ways. Therfore,favourable number of events = 29 C 2 * 20 C 2 Hence,required probability = 29 C 2 * 20 C 2 / 50 C 5 =551 / 15134

Question 4

A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident

Accepted Answer

Let A = Event that A speaks the truth B = Event that B speaks the truth Then P(A) = 75/100 = 3/4 P(B) = 80/100 = 4/5 P(A-lie) = 1 - 3 4 = 1/4 P(B-lie) = 1 - 4 5 = 1/5 Now, A and B contradict each other =[A lies and B true] or [B true and B lies] = P(A).P(B-lie) + P(A-lie).P(B) = 3 5 * 1 5 + 1 4 * 4 5 = 7 20 = 7 20 * 100 = 35%

Question 5

A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

Accepted Answer

A red ball can be drawn in two mutually exclusive ways (i) Selecting bag I and then drawing a red ball from it. (ii) Selecting bag II and then drawing a red ball from it. Let E1, E2 and A denote the events defined as follows: E1 = selecting bag I, E2 = selecting bag II A = drawing a red ball Since one of the two bags is selected randomly, therefore P(E1) = 1/2 and P(E2) = 1/2 Now, P A E 1 = Probability of drawing a red ball when the first bag has been selected = 4/7 P A E 2 = Probability of drawing a red ball when the second bag has been selected = 2/6 Using the law of total probability, we have P(red ball) = P(A) = P E 1 × P A E 1 + P E 2 × P A E 2 = 1 2 × 4 7 + 1 2 × 2 6 = 19 42

Question 6

There are 26 balls marked with alphabetical order A to Z. What is the probability of selecting vowels listed balls?

Accepted Answer

We know that, Total number of balls n(S) = 26 Number of vowels n(E) = 5 Hence, required probability = n(E)/n(S) = 5/26.

Question 7

A word consists of 9 letters; 5 consonants and 4 vowels.Three letters are choosen at random. What is the probability that more than one vowel will be selected ?

Accepted Answer

3 letters can be choosen out of 9 letters in 9 C 3 ways. More than one vowels ( 2 vowels + 1 consonant or 3 vowels ) can be choosen in ( 4 C 2 * 5 C 1 ) + 4 C 3 ways Hence,required probability = 4 C 2 * 5 C 1 + 4 C 3 9 C 3 = 17/42

Question 8

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :

Accepted Answer

One person can select one house out of 3= 3 C 1 ways =3. Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9. Therefore, probability that all thre apply for the same house is 1/9

Question 9

The probabilities that a student will receive an A, B, C or D grade are 0.4, 0.3 , 0.2 and 0.1 respectively. Find the probability that a student will receive Atleast B grade.

Accepted Answer

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7

Question 10

A box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains exactly one defective bulb.

Accepted Answer

Total number of elementary events = 10 C 5 Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is 3 C 1 * 7 C 4 So,required probability = 3 C 1 * 7 C 4 / 10 C 5 = 5/12.

Question 11

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

Accepted Answer

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36. Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3,4),(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} n(E) = 27. P(E) = n(E)/n(S) = 27/36 = 3/4.

Question 12

What is the probability of getting a sum 9 from two throws of a dice?

Accepted Answer

In two throws of a die, n(S) = (6 x 6) = 36. Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}. P(E) =n(E)/n(S)=4/36=1/9.

Question 13

In a single throw of two dice , find the probability that neither a doublet nor a total of 8 will appear.

Accepted Answer

n(S) = 36 A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) } n A = 6 , n B = 5 , n A ∩ B = 1 ∴ Required probability = P A ∪ B = P A + P B - P A ∩ B = 6 36 + 5 36 - 1 36 = 5 18

Question 14

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Accepted Answer

Total number of balls = (8 + 7 + 6) = 21. Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue. n(E) = 7. P(E) = n(E)/n(S) = 7/21 = 1/3.

Question 15

Four dice are thrown simultaneously. Find the probability that all of them show the same face.

Accepted Answer

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is: = 6 * 6 * 6 * 6 = 6 4 n(S) = 6 4 Let X be the event that all dice show the same face. X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)} n(X) = 6 Hence required probability = n ( X ) n ( S ) = 6 6 4 = 1 216

Question 16

The probability that z lies between 0 and 3.01

Accepted Answer

Probability between z = 0 and z = 3.01 is given by P(0<z<3.01) = P(z<3.01) - P(z<0) Reading from the z-table, we have P(z<0) = 0.5 P(z<3.01) = 0.9987 Hence, P(0<z<3.01) = 0.9987 - 0.5 = 0.4987.

Question 17

What is the probability that a number is divisible by 4?

Accepted Answer

Probability is Number of favourable outcomes by total outcomes Total comes = 2 (Divisible or not) Favourable = 1 (Divisible) Hence, probability = 1/2.

Question 18

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?

Accepted Answer

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.

Question 19

In a simultaneous throw of pair of dice. Find the probability of getting the total more than 7.

Accepted Answer

Here n(S) = (6 x 6) = 36 Let E = event of getting a total more than 7 = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)} Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.

Question 20

A number X is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that |X|<2

Accepted Answer

X can take 7 values. To get |X|+2) take X={−1,0,1} => P(|X|<2) = Favourable CasesTotal Cases = 3/7

Probability Questions