100+ Probability TNPSC Previous Year Questions - 1
Question: 1
Three unbiased coins are tossed. What is the probability of getting at least 2 heads?
(A) $${1}/{2}$$
(B) $${1}/{3}$$
(C) $${1}/{4}$$
(D) $${1}/{8}$$
Ans: A
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.
∴ P (E) = $${n(E)}/{n(S)}$$ = $${4}/{8}$$ = $${1}/{2}$$.
Question: 2
Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
(A) $${1}/{2}$$
(B) $${3}/{4}$$
(C) $${3}/{8}$$
(D) $${5}/{16}$$
Ans: B
In a simultaneously throw of two dice, we have n(S) = (6 × 6) = 36.
Let E = event of getting two numbers whose product is even.
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: 3
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?
(A) $${2}/{3}$$
(B) $${3}/{4}$$
(C) $${8}/{21}$$
(D) $${9}/{23}$$
Ans: C
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 red.
∴ n(E) = 8.
∴ P(E) = $${8}/{21}$$.
Question: 4
An urn contains 3 red, 3 green and 2 blue balls. If 2 balls are drawn at random, find the probability that no ball is blue.
(A) $${2}/{7}$$
(B) $${5}/{7}$$
(C) $${10}/{21}$$
(D) $${11}/{21}$$
Ans: C
Total number of balls = (2 + 3 + 2) = 7.
Let E be the event of drawing 2 non blue balls.
Then, n(E) = 5C2 = $${5 × 4} / {2 × 1}$$ = 10.
Also, n(S) = 7C2 = $${7 × 6} / {2 × 1}$$ = 21.
∴ P(E) = $${n(E)} / {n(S)}$$ = $${10}/{21}$$.
Question: 5
Two dice are tossed. The probability that the total score is a prime number is
(A) $${1}/{2}$$
(B) $${1}/{6}$$
(C) $${5}/{12}$$
(D) $${7}/{9}$$
Ans: C
Clearly, n(S) = (6 × 6) = 36.
Let E be the event that the sum is a prime number.
Then, n(E) = {(1, 1), (1, 2), (1, 4), (1, 6),(2, 1), (2, 3), (2, 5), (3, 2),(3, 4), (4, 1), (4, 3), (5, 2),(5, 6), (6, 1), (6, 5) }
∴ n(E) = 15.
∴ P(E) = $${n(E)}/{n(S)}$$ = $${15}/{36}$$ = $${5}/{12}$$.
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