Probability — Part 2 MCQs with Explanations

Let $${E_1}$$ = event that A speaks the truth And $${E_2}$$ = event that B speaks the truth Then, $$\eqalign{ & P\left( {{E_1}} \right) = \frac{{60}}{{100}} = \frac{3}{5}, \cr & P\left( {{E_2}} \right) = \frac{{70}}{{100}} = \frac{7}{{10}}, \cr & P\left( {{{\bar E}_1}} \right) = \left( {1 - \frac{3}{5}} \right) = \frac{2}{5}, \cr & P\left( {{{\bar E}_2}} \right) = \left( {1 - \frac{7}{{10}}} \right) = \frac{3}{{10}} \cr} $$ P (A and B say the same thing) = P [(A speaks the truth and B speaks the truth) or (A tells a lie and B tells a lie)] $$=$$ P [$$\left( {{E_1} \cap {E_2}} \right)$$ or $$({\overline E _1} \cap {\overline E _2})] $$ $$=$$ P $$\left( {{E_1} \cap {E_2}} \right)$$ + $$({\overline E _1} \cap {\overline E _2})$$ $$=$$ P ($${{E_1}}$$). P ($${{E_2}}$$) + P ($${\overline E _1}$$) . P ($${\overline E _2}$$) $$\eqalign{ & = \left( {\frac{3}{5} \times \frac{7}{{10}}} \right) + \left( {\frac{2}{5} \times \frac{3}{{10}}} \right) \cr & = \frac{{27}}{{50}} \cr & = 0.54 \cr} $$

Let S be the sample space and let E be the event of selecting 2 boys and 1 girl. Then, n(S) = number of ways of selecting 3 students out of 25 = $${}^{25}\mathop C\nolimits_3 = $$ $$\frac{{25 \times 24 \times 23}}{{3 \times 2 \times 1}}$$ = 2300 And, n(E) = $$\left( {{}^{15}\mathop C\nolimits_2 \times {}^{10}\mathop C\nolimits_1 } \right)$$ $$ = \left( {\frac{{15 \times 14}}{{2 \times 1}} \times 10} \right)$$ = 1050 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}}$$

Total number of balls = (6 + 2 + 4 + 3) = 15 Let E be the event of drawing 4 balls such that 2 are red and 2 are green. Then, n(E) = $$\left( {{}^2\mathop C\nolimits_2 \times {}^4\mathop C\nolimits_2 } \right)$$ $$ = \left( {1 \times \frac{{4 \times 3}}{{2 \times 1}}} \right)$$ = 6 And, n(S) = $${}^{15}\mathop C\nolimits_4 = $$ $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$ = 1365 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{1365}} = \frac{2}{{455}}$$

$$\eqalign{ & P(E) = \frac{{30}}{{100}} = \frac{3}{{10}}; \cr & P(H) = \frac{{20}}{{100}} = \frac{1}{5}\,\,{\text{and}} \cr & {\text{P }}\left( {{\text{E}} \cap {\text{H}}} \right)\,{\text{ = }}\frac{{10}}{{100}} = \frac{1}{{10}} \cr} $$ P (E or H) = $$P(E \cup H)$$ = P(E) + P(H) - $$P(E \cup H)$$ $$\eqalign{ & = \left( {\frac{3}{{10}} + \frac{1}{5} - \frac{1}{{10}}} \right) \cr & = \frac{4}{{10}} \cr & = \frac{2}{5} \cr} $$

Total number of marbles = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 2 blue and 1 yellow marble. Then, n(E) = $$\left( {{}^4\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right)$$ $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times 3} \right)$$ = 18 Also, n(S) = $${}^{15}\mathop C\nolimits_3 = $$ $$\frac{{15 \times 14 \times 13}}{{3 \times 2 \times 1}}$$ = 455 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{18}}{{455}}$$

n(S) = number of ways of choosing 4 persons out of 9 $$ = {}^9\mathop C\nolimits_4 $$ $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$ = 126 n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal n(E) $$ = \left( {{}^4\mathop C\nolimits_2 \times {}^5\mathop C\nolimits_2 } \right)$$ $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$ = 60 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{60}}{{126}} = \frac{{10}}{{21}}$$

Total number of balls = (6 + 2 + 4 + 3) = 15 Let E be the event of drawing 3 non-yellow balls Then, n(E) = $${}^{12}\mathop C\nolimits_3 $$ $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$ = 220 Also, n(S) = $${}^{15}\mathop C\nolimits_3 $$ $$ = \frac{{15 \times 14 \times 13}}{{3 \times 2 \times 1}}$$ = 455 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{220}}{{455}} = \frac{{44}}{{91}}$$

Here, n(S) = 52 There are 13 cards of diamond (including one king) and there are three more kings. Let E = event of getting a diamond or a king Then, n(E) = (13 + 3) = 16 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{16}}{{52}} = \frac{4}{{13}}$$

Total number of persons = (3 + 2) = 5 $$\therefore n(S) = {}^5\mathop C\nolimits_3 = {}^5\mathop C\nolimits_2 $$ $$ = \frac{{5 \times 4}}{{2 \times 1}}$$ = 10 Let E be the event of selecting 3 members having at least 1 women Then, n(E) = n [(1 women and 2 men ) or (2 women and 1 man)] = n (1 woman and 2 men) + n (2 women and 1 man) $$\eqalign{ & = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_2 } \right) + \left( {{}^2\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right) \cr & = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_1 } \right) + \left( {1 \times {}^3\mathop C\nolimits_1 } \right) \cr & = \left( {2 \times 3} \right) + \left( {1 \times 3} \right) \cr & = \left( {6 + 3} \right) \cr & = 9 \cr} $$ $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{9}{{10}}$$

Let $${{{\text{E}}_1}}$$ = event that A speaks the truth And $${{{\text{E}}_2}}$$ = event that B speaks the truth Then, $$\eqalign{ & P\left( {{E_1}} \right) = \frac{{75}}{{100}} = \frac{3}{4}, \cr & P\left( {{E_2}} \right) = \frac{{80}}{{100}} = \frac{4}{5}, \cr & P\left( {{{\overline E }_1}} \right) = \left( {1 - \frac{3}{4}} \right) = \frac{1}{4}, \cr & P\left( {{{\overline E }_2}} \right) = \left( {1 - \frac{4}{5}} \right) = \frac{1}{5} \cr} $$ P (A and B contradict each other) = P [(A speaks the truth and B tells a lie) or (A tells a lie and B speaks the truth)] $$\eqalign{ & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right)or\left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right) + \left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right) + P\left( {{{\overline E }_1}} \right).P\left( {{E_2}} \right) \cr & = \left( {\frac{3}{4} \times \frac{1}{5}} \right) + \left( {\frac{1}{4} \times \frac{4}{5}} \right) \cr & = \left( {\frac{3}{{20}} + \frac{1}{5}} \right) \cr & = \frac{7}{{20}} \cr & = \left( {\frac{7}{{20}} \times 100} \right)\% \cr & = 35\% \cr} $$

Total number of balls = (4 + 5 + 3) = 12 Let E be the event of drawing 2 red balls. Then, n (E) = $${}^4\mathop C\nolimits_2 = \frac{{4 \times 3}}{{2 \times 1}}$$ = 6 Also n (S) = $${}^{12}\mathop C\nolimits_2 = \frac{{12 \times 11}}{{2 \times 1}}$$ = 66 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{66}} = \frac{1}{{11}}$$

Total number of balls = (8 + 7 + 6) = 21 Let E = event that the ball drawn is neither blue nor green = Even that the ball drawn is red ∴ n(E) = 8 P(E) = $$\frac{{8}}{{21}}$$

Probability of hitting the target: Dev can hit target ⇒ $$\frac{{3}}{{6}}$$ =$$\frac{{1}}{{2}}$$ Lakhan can hit target =$$\frac{{4}}{{4}}$$ = 1 Pawan can hit target = $$\frac{{2}}{{6}}$$ = $$\frac{{1}}{{3}}$$ Required probability that at least 2 shorts hit target $$\eqalign{ & = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} \cr & = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} \cr & = \frac{4}{6} \cr & = \frac{2}{3} \cr} $$

Number of red balls = 4 Number of yellow balls = 5 Number of pink balls = 6 Total balls = 4 + 5 + 6 = 15 Total possible outcomes = selection of 2 balls out of 15 balls = $${}^{15}\mathop C\nolimits_2 $$ =$$\frac{{15!}}{{2!(15 - 2)!}}$$ =$$\frac{{15!}}{{2! × 13!}}$$ =$$\frac{{15 × 14}}{{1 × 2}}$$ =105 Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls. $$ = {}^{10}\mathop C\nolimits_2 $$ = $$\frac{{10!}}{{2!(10 - 2)!}}$$ = $$\frac{{10!}}{{2! × 8!}}$$ = $$\frac{{10 × 9}}{{1 × 2}}$$ = 45 ∴ Required probability = $$\frac{{45}}{{105}}$$ = $$\frac{{3}}{{7}}$$

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) = $$\frac{{n (E)}}{{n (S)}}$$ = $$\frac{{6}}{{36}}$$ = $$\frac{{1}}{{6}}$$

P(Both are red), $$\eqalign{ & = \frac{{^6{C_2}}}{{^{13}{C_2}}} \cr & = \frac{5}{{26}} \cr} $$

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, $$\eqalign{ & = \frac{1}{2} \times \frac{1}{6} \cr & = \frac{1}{{12}} \cr} $$

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: $$\eqalign{ & = P(A) + P(B) \cr & = \frac{{^{12}{C_1}{ \times ^1}{C_1}}}{{^{52}{C_2}}} + \frac{{^{13}{C_1}{ \times ^3}{C_1}}}{{^{52}{C_2}}} \cr & = \left( {\frac{{2 \times \left( {12 \times 1} \right)}}{{52 \times 51}}} \right) + \left( {\frac{{2\left( {13 \times 3} \right)}}{{52 \times 51}}} \right) \cr & = \left( {\frac{{24 + 78}}{{52 \times 51}}} \right) \cr & = \frac{1}{{26}} \cr} $$

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 $$\eqalign{ & P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( S \right)}} \cr & = \frac{{10!\, \times 2!}}{{11!}} \cr & = \frac{2}{{11}} \cr} $$

Let the probability of winning the race is denoted by P(person) $$\eqalign{ & P(P) = \frac{1}{4}, \cr & P(Q) = \frac{1}{5}, \cr & P(R) = \frac{1}{6}, \cr & P(S) = \frac{1}{7} \cr} $$ All the events are mutually exclusive (since if one of them wins then other would lose as pointed out by rahul) hence, Required probability: $$ = P(P) + P(Q) + P(R)$$ $$ + P(S)$$ $$\eqalign{ & = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \cr & = \frac{{319}}{{420}} \cr} $$

There are 13 spade and 3 more jack Probability of getting spade or a jack: $$\eqalign{ & = \frac{{13 + 3}}{{52}} \cr & = \frac{4}{{13}} \cr} $$ So probability of getting neither spade nor a jack: $$\eqalign{ & = 1 - \frac{4}{{13}} \cr & = \frac{9}{{13}} \cr} $$

X = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} n(X) = 10, n(S) = 30 Hence required probability, $$\eqalign{ & = \frac{{n(X)}}{{n(S)}} \cr & = \frac{{10}}{{30}} \cr} $$

250 numbers between 101 and 350 i.e. n(S) = 250 n(E) = 100 th digits of 2 = 299 - 199 = 100 $$\eqalign{ & P(E) = \frac{{n(E)}}{{n(S)}} \cr & = \frac{{100}}{{250}} \cr & = 0.40 \cr} $$

Required probability: $$\eqalign{ & = \frac{{^4{C_1}{ \times ^4}{C_1}{ \times ^4}{C_1}}}{{^{52}{C_3}}} \cr & = \frac{{4 \times 4 \times 4}}{{22100}} \cr & = \frac{{16}}{{5525}} \cr} $$

The probability that exactly 4 vessels arrive safely is, $${ = ^5}{C_4} \times {\left( {\frac{9}{{10}}} \right)^4}\left( {\frac{1}{{10}}} \right)$$ The probability that all 5 arrive safely is $${\left( {\frac{9}{{10}}} \right)^5}$$ The probability that at-least 4 vessels arrive safely, $$ = {}^5{C_4} \times {\left( {\frac{9}{{10}}} \right)^4}\left( {\frac{1}{{10}}} \right) + $$ $${\left( {\frac{9}{{10}}} \right)^5}$$ $$ = \frac{{14 \times {9^4}}}{{{{10}^5}}}$$

The total number of words which can be formed by permuting the letters of the word 'UNIVERSITY' is $$\frac{{10!}}{{2!}}$$ as there is two I's. Hence $$n(S) = \frac{{10!}}{{2!}}$$ Taking two I's as one letter, number of ways of arrangement in which both I's are together = 9! $${\text{So}}\,n\left( X \right) = 9!$$ Hence required probability $$\eqalign{ & = \frac{{n(X)}}{{n(S)}} \cr & = \frac{{9!}}{{10!/2!}} \cr & = \frac{1}{5} \cr} $$

A number divisible by 4 formed using the digits 1, 2, 3, 4 and 5 has to have the last two digits 12 or 24 or 32 or 52. In each of these cases, the five digits number can be formed using the remaining 3 digits in 3! = 6 ways. A number divisible by 4 can be formed in 6 × 4 = 24 ways. Total number that can be formed using the digits 1, 2, 3, 4 and 5 without repetition = 5! = 120 Required probability, $$\eqalign{ & = \frac{{24}}{{120}} \cr & = \frac{1}{5} \cr} $$

P(None is blue), $$\eqalign{ & = \frac{{^8{C_3}}}{{^{13}{C_3}}} \cr & = \frac{{56}}{{286}} \cr & = \frac{{28}}{{143}} \cr} $$ P(At least one is blue), $$\eqalign{ & = 1 - \frac{{28}}{{143}} \cr & = \frac{{115}}{{143}} \cr} $$

$$\eqalign{ & {\text{Here}} \cr & n\left( E \right){ = ^7}{C_1}{ \times ^5}{C_1}{ \times ^5}{C_1} \cr & {\text{and,}} \cr & n\left( S \right){ = ^{12}}{C_1}{ \times ^{11}}{C_1}{ \times ^{10}}{C_1} \cr & P(S) = \frac{{7 \times 6 \times 5}}{{12 \times 11 \times 10}} \cr & = \frac{7}{{44}} \cr} $$

As probability of a both the teams (Arrogant and Overconfident) winning simultaneously is zero. $$\eqalign{ & P\left( {A \cap O} \right) = 0 \cr & P\left( {A \cap B} \right) = P\left( A \right) + P\left( B \right) \cr & = \frac{5}{8} + \frac{1}{5} \cr & = \frac{{33}}{{40}} \cr} $$ So required odds will be 33 : 7

Given, A be the event that X is selected and B is the event that Y is selected. $$P(A) = \frac{1}{7},\,P(B) = \frac{2}{9}$$ Let C be the event that both are selected. P(C) = P(A) × P(B) as A and B are independent events: $$\eqalign{ & = \left( {\frac{1}{7}} \right) \times \left( {\frac{2}{9}} \right) \cr & = \frac{2}{{63}} \cr} $$

Four cards can be selected from 52 cards in $$^{52}{C_4}$$ ways. Now, there are four suits, e.g. club, spade, heart and diamond each of 13 cards. So total number of ways of getting all the four cards of the same suit: $$\eqalign{ & { \Rightarrow ^{13}}{C_4}{ + ^{13}}{C_4}{ + ^{13}}{C_4}{ + ^{13}}{C_4} \cr & = 4{ \times ^{13}}{C_4} \cr} $$ So required probability, $$\eqalign{ & = \frac{{4{ \times ^{13}}{C_4}}}{{^{52}{C_4}}} \cr & = \frac{{44}}{{4165}} \cr} $$

The probability that first toy shows the even number $$ = \frac{{10}}{{21}}$$ Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left. Hence, probability that second toy shows the even number $$ = \frac{{9}}{{20}}$$ Required probability, $$\eqalign{ & = \left( {\frac{{10}}{{21}}} \right) \times \left( {\frac{9}{{20}}} \right) \cr & = \frac{9}{{42}} \cr} $$

Let X be required events and S be the sample space then X = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (6, 2), (3, 4), (6, 4), (3, 6)} n(X) = 11, n(S) = 36 Hence, required probability $$\eqalign{ & = \frac{{n(X)}}{{n(S)}} \cr & = \frac{{11}}{{36}} \cr} $$

$$P(X) = \frac{1}{5},$$ $$P(Y) = \frac{1}{4},$$ $$P(Z) = \frac{1}{3}$$ Required probability: $$ = \left[ {P\left( A \right)P\left( B \right)\left\{ {1 - P\left( C \right)} \right\}} \right]$$ $$ + $$ $$\left[ {\left\{ {1 - P\left( A \right)} \right\}P\left( B \right)P\left( C \right)} \right] + $$ $$\left[ {P\left( A \right)P\left( C \right)\left\{ {1 - P\left( B \right)} \right\}} \right] + $$ $$\left[ {P\left( A \right)P\left( B \right)P\left( C \right)} \right]$$ $$ = \left[ {\frac{1}{4} \times \frac{1}{3} \times \frac{4}{5}} \right] + $$ $$\left[ {\frac{3}{4} \times \frac{1}{3} \times \frac{1}{5}} \right] + $$ $$\left[ {\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}} \right] + $$ $$\left[ {\frac{1}{4} \times \frac{1}{3} \times \frac{1}{5}} \right]$$ $$\eqalign{ & = \frac{4}{{60}} + \frac{3}{{60}} + \frac{2}{{60}} + \frac{1}{{60}} \cr & = \frac{{10}}{{60}} \cr & = \frac{1}{6} \cr} $$

The probability that first ball is white: $$\eqalign{ & = \frac{{^{12}{C_1}}}{{^{30}{C_1}}} \cr & = \frac{{12}}{{30}} \cr & = \frac{2}{5} \cr} $$ Since, the ball is not replaced; hence the number of balls left in bag is 29. Hence, the probability the second ball is black: $$\eqalign{ & = \frac{{^{18}{C_1}}}{{^{29}{C_1}}} \cr & = \frac{{18}}{{29}} \cr} $$ Required probability, $$\eqalign{ & = \left( {\frac{2}{5}} \right) \times \left( {\frac{{18}}{{29}}} \right) \cr & = \frac{{36}}{{145}} \cr} $$

Total cases of checking in the hotels = $${4^3}$$ ways. Cases, when 3 men are checking in different hotels = 4 × 3 × 2 = 24 ways. Required probability: $$\eqalign{ & = \frac{{24}}{{{4^3}}} \cr & = \frac{3}{8} \cr} $$

Different possible cases of contradiction, A speaks truth and B does not speaks truth. Or, A does not speak truth and B speaks truth. $$\eqalign{ & = \left( {\frac{3}{4} \times \frac{1}{5}} \right) + \left( {\frac{1}{4} \times \frac{4}{5}} \right) \cr & = \frac{3}{{20}} + \frac{4}{{20}} \cr & = \frac{7}{{20}} \cr & = 35\% \cr} $$

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is: $$\eqalign{ & = 6 \times 6 \times 6 \times 6 = {6^4} \cr & n(S) = {6^4} \cr} $$ Let X be the event that all dice show the same face. X = {(1, 1, 1, 1), (2, 2, 2, 2), (3, 3, 3, 3), (4, 4, 4, 4), (5, 5, 5, 5), (6, 6, 6, 6)} n(X) = 6 Hence required probability, $$\eqalign{ & = \frac{{n(X)}}{{n(S)}} = \frac{6}{{{6^4}}} \cr & = \frac{1}{{216}} \cr} $$

Let probability of the first event taking place be A and probability of the second event taking place be B. Then $$\eqalign{ & P\left( A \right) = \frac{3}{{5 + 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{8} \cr & P\left( B \right) = \frac{7}{{7 + 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{7}{{12}} \cr} $$ The required event can be defined as that A takes place and B does not take place (A or B takes place and A does not take place or A takes place and B takes place.) $$ = \left[ {P\left( A \right)\left\{ {1 - P\left( B \right)} \right\}} \right] + $$ $$\left[ {P\left( B \right)\left\{ {1 - P\left( A \right)} \right\}} \right] + $$ $$\left[ {P\left( A \right)P\left( B \right)} \right]$$ $$ = \left[ {\frac{3}{8} \times \frac{5}{{12}}} \right] + $$ $$\left[ {\frac{5}{8} \times \frac{7}{{12}}} \right] + $$ $$\left[ {\frac{3}{8} \times \frac{7}{{12}}} \right]$$ $$\eqalign{ & = \frac{{15 + 35 + 21}}{{96}} \cr & = \frac{{71}}{{96}} \cr} $$