Probability - Study Mode
[#71] In a simultaneous throw of two dice, what is the probability of getting a doublet ?
Correct Answer
(A) $$frac{{1}}{{6}}$$
Explanation
Solution: In a simultaneous throw of dice, n (S) = (6 × 6) = 36 Let E = event of getting a doublet = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] ∴ P(E) = $$frac{{n (E)}}{{n (S)}}$$ = $$frac{{6}}{{36}}$$ = $$frac{{1}}{{6}}$$
[#72] An urn contains 6 red, 5 blue and 2 green marbles. If 2 marbles are picked at random, what is the probability that both are red?
Correct Answer
(B) $$frac{{5}}{{26}}$$
Explanation
Solution: P(Both are red), $$eqalign{
& = frac{{^6{C_2}}}{{^{13}{C_2}}} cr
& = frac{5}{{26}} cr} $$
[#73] You toss a coin AND roll a die. What is the probability of getting a tail and a 4 on the die?
Correct Answer
(D) $$frac{{1}}{{12}}$$
Explanation
Solution: Probability of getting a tail when a single coin is tossed $$ = frac{1}{2}$$ Probability of getting 4 when a die is thrown $$ = frac{1}{6}$$ Required probability, $$eqalign{
& = frac{1}{2} imes frac{1}{6} cr
& = frac{1}{{12}} cr} $$
[#74] Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King.
Correct Answer
(D) $$frac{{1}}{{26}}$$
Explanation
Solution: Let X be the event that cards are in a club which is not king and other is the king of club. Let Y be the event that one is any club card and other is a non-club king. Hence, required probability: $$eqalign{
& = P(A) + P(B) cr
& = frac{{^{12}{C_1}{ imes ^1}{C_1}}}{{^{52}{C_2}}} + frac{{^{13}{C_1}{ imes ^3}{C_1}}}{{^{52}{C_2}}} cr
& = left( {frac{{2 imes left( {12 imes 1}
ight)}}{{52 imes 51}}}
ight) + left( {frac{{2left( {13 imes 3}
ight)}}{{52 imes 51}}}
ight) cr
& = left( {frac{{24 + 78}}{{52 imes 51}}}
ight) cr
& = frac{1}{{26}} cr} $$
[#75] P and Q sit in a ring arrangement with 10 persons. What is the probability that P and Q will sit together?
Correct Answer
(A) $$frac{{2}}{{11}}$$
Explanation
Solution: n(S)= number of ways of sitting 12 persons at round table: = (12 - 1)! = 11! Since two persons will be always together, then number of persons: = 10 + 1 = 11 So, 11 persons will be seated in (11 - 1)! = 10! ways at round table and 2 particular persons will be seated in 2! ways. n(A) = The number of ways in which two persons always sit together = 10! × 2 $$eqalign{
& Pleft( A
ight) = frac{{nleft( A
ight)}}{{nleft( S
ight)}} cr
& = frac{{10!, imes 2!}}{{11!}} cr
& = frac{2}{{11}} cr} $$