Practice Model question set 1 - 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) = ${n(E)}/{n(S)}$ = ${3}/{12}$ = $1/4$