Practice Probability - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A student has 60% chance of passing in English and 54% chance of passing in both English and Mathematics. What is the percentage probability that he will fail in Mathematics?
(a)
(b)
(c)
(d)
(e)
P(E) = Probability of passing in English = 0.6
P(E ∩ M) = Probability of passing in Maths and English
= 0.54
P(M) = Probability of passing in Maths
Since, P(M) and P(E), both are independent events.
So, P(E ∩ M) = P(E) × P(M)
P(M) = P(E ∩ M)/ P(E) = ${0.54}/{0.6}$ = 0.9
∴ Probability of failing in Maths = 1 – 0.9 = 0.1 = 10%
Q-2) Each of the 3 persons is to be given some identical items such that product of the numbers of items received by each of the three persons is equal to 30. In how many maximum different ways can this distribution be done?
(a)
(b)
(c)
(d)
(e)
Suppose three people have been given a, b and c number of items.
Then, a × b × c = 30
Now, There can be 5 cases :
Case I : When one of them is given 30 items and rest two 1 item each.
So, number of ways for (30 × 1 × 1) = ${3!}/{2!}$ = 3
(As two of them have same number of items)
Case II : Similarly, number of ways for (10 × 3 × 1) = 3! = 6
Case III : Number of ways for (15 × 2 × 1) = 3! = 6
Case IV : Number of ways for (6 × 5 × 1) = 3! = 6
Case V : Number of ways for (5 × 3 × 2) = 3! = 6
Here, either of these 5 cases are possible.
Hence, total number of ways = 3 + 6 + 6 + 6 + 6 = 27
Q-3) 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)
(b)
(c)
(d)
(e)
Probability to be a Blue = ${^3C_3}/{^7C_3}$
Probability to be a Red = ${^4 C_3}/{^7C_3}$
Required probability = ${^3 C_3}/{^7 C_3}$ + ${^4 C_3}/{^7C_3}$ = $2/{35}$
Q-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)
(b)
(c)
(d)
(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 = ${n(E)}/{n(S)}$ = $6/{66}$ = $1/{11}$
Q-5) A dice is thrown. What is the probability that the number shown on the dice is not divisible by 3?
(a)
(b)
(c)
(d)
(e)
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) = $4/6$ = $2/3$
Q-6) When three coins are tossed together, the probability that all coins have the same face up, is
(a)
(b)
(c)
(d)
(e)
Probability of Head or Tail on the upper side for a coin
= $1/2$ ∴ Probability of same side on the upper side for the three coins = ${1/2} × {1/2} × {1/2}$ = $(1/2)^3$ = $1/8$
∴ angles are 80°, 60° and 40°
Q-7) A box contains 4 black balls, 3 red balls and 5 green balls. 2 balls are drawn from the box at random. What is the probability that both the balls are of the same colour?
(a)
(b)
(c)
(d)
(e)
Total no. of balls =4 + 3 + 5 = 12
n(S) = $^{12}C_2$ = ${12 × 11}/2$ = 66
n(E) = $ ^4C_2+ ^3C_2 + ^5C_2 $ = ${{4 × 3}/2} + {{3 × 2}/2} + {{5 × 4}/2}$
= 6 + 3 + 10 = 19
∴ Required probability, P(E) = ${n(E)}/{n(S)}$ = ${19}/{66}$
Q-8) Ten identical particles are moving randomly inside a closed box. What is the probability that at any given point of time all the ten particles will be lying in the same half of the box?
(a)
(b)
(c)
(d)
(e)
Probability of a particle lying in any particular half = $1/2$
∴ Probability of all 10 particles lying in either 1st half or
2nd half = $(1/2)^{10} + (1/2)^{10}$ = 2$(1/2)^{10}$ = $2^{1/9}$
Q-9) A committee of 4 is to be formed from among 4 girls and 5 boys. What is the probability that the committee will have number of boys less than number of girls?
(a)
(b)
(c)
(d)
(e)
Selection of 1 boy and 3 girls in $^5 C_1$ × $^4 C _3$ = 5 × 4 = 20 ways
Selection of 4 girls and no boy in $^5 C_0$ × $^4 C _4$ = 1 × 1 = 1 way
∴ n(E) = total no. of ways = 21
Without any restriction, a committee of 4 can be formed from among 4 girls and 5 boys in $^9C_4$
= ${9 × 8 × 7 × 6}/{4 × 3 × 2}$ = 9 × 7 × 2 ways
∴ P(E) = ${n(E)}/{n(S)}$ = ${21}/{9 × 7 × 2}$ = $1/6$
Q-10) Directions :
Study the given information carefully and answer the questions that follow.
A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap.
If one cap is picked at random, what is the probability that it is either blue or yellow?
(a)
(b)
(c)
(d)
(e)
Total number of caps = 12
n (S) = $^{12}C_1$ = 12
Out of (2 blue + 1 yellow) caps number of ways to pick one cap n(E) = $^3C_1$ = 3
Required probability p(E) = ${n(E)}/{n(S)}$ = ${3}/{12}$ = $1/4$
Q-11) 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)
(b)
(c)
(d)
(e)
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.
= ${8}/{104}$ × ${7}/{103}$ = $7/{1339}$
Q-12) The letters of the word ‘ARTICLE’ are arranged in different ways randomly. What is the chance that the vowels occupy the even places?
(a)
(b)
(c)
(d)
(e)
The 7 different letters of the word ARTICLE can be arranged in 7! ways, i.e., n(S) = 7!
n(E) = $^3 P_3$ × $^4P_4$ = 3! × 4! = 6 × 34
∴ P(E) = ${6 × 24}/{7!}$ = $1/{35}$
Q-13) Directions :
Study the given information carefully and answer the questions that follow.
A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap.
If four caps are picked at random, what is the probability that none is green ?
(a)
(b)
(c)
(d)
(e)
Total number of caps = 12
Total result n(S) = ${12} C_4$.
n(S) = ${12!}/{4! × 12! - 4}$ = ${12 × 11 × 10 × 9 × 8!}/{4 × 3 × 2 × 1 × 8!}$ = 5 × 99
n$(E_1)$ = Out of 5 caps, number of ways to not pick a green cap = $^5C_0$.
n$(E_2)$ = Out of 7 caps, number of ways to pick 4 caps
= $^7C_4$. = ${7!}/{4! × 7! - 4}$ = ${7 × 6 × 5 × 4 × 3!}/{4 × 3 × 2 × 1 × 3!}$ = 35
p(E) = ${n(E_1) n (E_2)}/{n(s)}$ = ${1 × 35}/{5 × 99}$ = ${7}/{99}$
Q-14) Directions :
Study the given information carefully and answer the questions that follow.
A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap.
If three caps are picked at random, what is the probability that two are red and one is green ?
(a)
(b)
(c)
(d)
(e)
Total number of caps = 12
n(S) = $^{12}C_3$ = ${12!}/{3! × 12! -3}$ = ${12 × 11 × 10 × 9!}/{3 × 2 × 1 × 9!}$ = 220
n$(E_1)$ = Out of 4 red caps, number of ways to pick 2 caps = $^4C_2$
= ${4!}/{2! × 4! -2}$ = ${4 × 3 × 2 × 1}/{2 × 1 × 2 × 1}$ = 6.
n$(E_2)$ = Out of 5 green caps, number of ways to pick one
cap = $^5C_1$ = 5
p(E) = ${n(E_1) × n(E_2)}/{n(S)}$ = ${6 × 5}/{220}$ = ${3}/{22}$
Q-15) Directions :
Study the given information carefully and answer the questions that follow.
A basket contains 4 red, 5 blue and 3 green marbles.
If three marbles are picked at random, what is the probability that either all are green or all are red ?
(a)
(b)
(c)
(d)
(e)
Total possible result = Selection of 3 marbles out of 12
= $^{12}C_3$ = ${12 × 11 × 10}/{1 × 2 × 3}$ = 220
Total number of event n(E) = $^3C_3 + ^4C_3$.
n (E) = 1 + 4 = 5
Required probability = ${n(E)}/{n(S)}$ = ${5}/{220}$ = ${1}/{44}$
Q-16) Directions :
Study the given information carefully and answer the questions that follow.
A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap.
If two caps are picked at random, what is the probability that both are blue ?
(a)
(b)
(c)
(d)
(e)
Total number of caps = 2 + 4 + 5 + 1 = 12
Total result n(S) = $^{12} C_ 2$.
n(S) = ${12!}/{2! × 12! - 2}$ = ${12!}/{2! × 10!}$ = ${12 × 11 × 10!}/{2 × 1 × 10!}$ = 66
Favorable result n(E) = $^2 C _2$ = 1
Required probability p(E) = ${n(E)}/{n(S)}$ = $1/{66}$
Q-17) 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)
(b)
(c)
(d)
(e)
P(At least one good) = 1 – P(All bad)
= 1 - ${{^4C_3}/{^12C_3}}$ = 1 - ${4/{220}}$ = 1 -${1/{55}}$= ${54}/{55}$
Q-18) Directions :
Study the given information carefully and answer the questions that follow.
A basket contains 4 red, 5 blue and 3 green marbles.
If three marbles are picked at random, what is the probability that at least one is blue ?
(a)
(b)
(c)
(d)
(e)
Total possible result = n(S) = $^{12} C_3$.
= ${12 × 11 × 10}/{1 × 2 × 3}$ = 220 Total number of event = n(E)
Except blue marbles, selection of 3 marbles out of 7 marbles
= $^{7}C_3$ = ${7 × 6 × 5}/{1 × 2 × 3}$ = 35
∴ Required probability = $(1 -{35}/{220}) = (1 - {7}/{44})$ = ${37}/{44}$
Q-19) 3 digits are chosen at random from 1,2,3,4,5,6,7,8 and 9 without repeating any digit. What is the probability that their product is odd?
(a)
(b)
(c)
(d)
(e)
Let E be the event of selecting the three numbers such that their product is odd and S be the sample space.
For the product to be odd, 3 numbers chosen must be odd.
∴ n(E) = $^5C_3$
n(S) = $^9C_3$
∴ P (E) = ${n(E)}/{n(S)}$ = ${^5C_3}/{^9C_3}$ = ${5}/{42}$
Q-20) There are 4 boys and 4 girls. They sit in a row randomly. What is the chance that all the girls do not sit together?
(a)
(b)
(c)
(d)
(e)