# Probability Model Questions Set 1 Practice Questions Answers Test With Solutions & More Shortcuts

#### PROBABILITY PRACTICE TEST [1 - EXERCISES]

Question : 1

Two packs of cards are thoroughly mixed and shuffled and two cards are drawn at random, one after the other. What is the probability that both of them are jacks?

a) \$2/13\$

b) \$1/169\$

c) \$1/13\$

d) \$7/1339\$

e) \$1/179\$

Total number of cards = 104 = 2 × 52

and total number of jacks = 8 = 2 × 4

∴ Probability for the jack in first draw = \$8/{104}\$

and probability for the jack in second draw = \$7/{103}\$

Since both the events are independent events.

Hence the probability that both of them are jacks.

= \${8}/{104}\$ × \${7}/{103}\$ = \$7/{1339}\$

Question : 2

A basket contains three blue and four red balls. If three balls are drawn at random from the basket, what is the probability that all the three are either blue or red ?

a) \$1/7\$

b) \$3/28\$

c) 1

d) \$3/14\$

e) None of these

Probability to be a Blue = \${^3C_3}/{^7C_3}\$

Probability to be a Red = \${^4 C_3}/{^7C_3}\$

Required probability = \${^3 C_3}/{^7 C_3}\$ + \${^4 C_3}/{^7C_3}\$ = \$2/{35}\$

Question : 3

In a box carrying one dozen of oranges, one-third have become bad. If 3 oranges are taken out from the box at random, what is the probability that at least one orange out of the three oranges picked up is good?

a) \$54/55\$

b) \$3/55\$

c) \$1/55\$

d) \$45/55\$

e) None of these

P(At least one good) = 1 – P(All bad)

= 1 - \${{^4C_3}/{^12C_3}}\$ = 1 - \${4/{220}}\$ = 1 -\${1/{55}}\$= \${54}/{55}\$

Question : 4

Directions :
Study the given information carefully and answer the questions that follow.

A basket contains 4 red, 5 blue and 3 green marbles.

If two marbles are drawn at random, what is the probability that both are red ?

a) \$1/2\$

b) \$1/6\$

c) \$3/7\$

d) \$2/11\$

e) None of these

Total possible result n(S) = \$^{12}C_{2}\$ = \${12 × 11}/{1 × 2}\$ = 66

Total number of event = n(E) \$^4 C _2\$ = \${4 × 3}/{1 × 2}\$ = 6

∴ Required probability = \${n(E)}/{n(S)}\$ = \$6/{66}\$ = \$1/{11}\$

Question : 5

A dice is thrown. What is the probability that the number shown on the dice is not divisible by 3?

a) \$2/3\$

b) \$2/5\$

c) \$1/3\$

d) \$1/4\$

e) None of these

S = {1, 2, 3, 4, 5, 6}; n(S) = 6

E(not divisible by 3) = 1, 2, 4, 5}, n(E) = 4

∴ P(not divisible by 3) = \$4/6\$ = \$2/3\$

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