Probability Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

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The following question based on Probability topic of quantitative aptitude

Questions : 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) $1/179$

The correct answers to the above question in:

Answer: (d)

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.

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

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Read more model question set 1 Based Quantitative Aptitude Questions and Answers

Question : 1

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

Answer: (e)

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 : 2

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

Answer: (a)

P(At least one good) = 1 – P(All bad)

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

Question : 3

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

Answer: (e)

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 = ${n(E)}/{n(S)}$ = $6/{66}$ = $1/{11}$

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Probability Model Questions Set 1

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