Maths MCQ

There are 5 girls and 2 boys.

They can be arranged among themselves in ${ }^7 P _7=7$ ! ways.

$\therefore$ Girls can be arranged among themselves in ${ }^5 P _5=5$ ! ways.

No two boys should sit together.

Let girls be denoted by the letter G.

– G – G – G – G – G –

There are 6 places, marked by ‘-’ where boys can sit.

∴ Boys can be arranged in

$\begin{aligned} & { }^6 P _2=\frac{6 !}{(6-2) !} \\ & =\frac{6 \times 5 \times 4 !}{4 !} \\ & =30 \text { ways. }\end{aligned}$

$\therefore$ Required probability $=\frac{5 ! \times 30}{7 !}=\frac{5 ! \times 30}{7 \times 6 \times 5 !}=\frac{5}{7}$

A fair dice is tossed twice.

∴ n(S) = 36

A: Getting 4, 5, or 6 on the first toss and Getting 1, 2, 3, or 4 on the second toss.

∴ A = {(4, 1), (4, 2), (4, 3), (4, 4), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}

$\therefore P ( A )=\frac{n(A)}{n(S)}=\frac{12}{36}=\frac{1}{3}$

∴ Required answer = P(A) : P(A’) = 1 : 2

Let event A: Student knows the correct answer,

event A’: Student guesses the answer,

event B: Answer is correct.

$\therefore P(A)=\frac{2}{3}, P\left(A^{\prime}\right)=\frac{1}{3}, P\left(B / A^{\prime}\right)=\frac{1}{4}$

Clearly, P(B/A) = 1 Required probability = P(A/B)

$\begin{aligned} & =\frac{ P ( A ) \cdot P ( B / A )}{ P ( A ) \cdot P ( B / A )+ P \left( A ^{\prime}\right) P \left( B / A ^{\prime}\right)} \\ & =\frac{\frac{2}{3} \times 1}{\frac{2}{3} \times 1+\frac{1}{3} \times \frac{1}{4}} \\ & =\frac{8}{9}\end{aligned}$

Let event S1: First shelve is selected, event S2: Second shelve is selected, event P: Drawing a physics book.

$\therefore P \left( S _1\right)=\frac{1}{2}$ and $P \left( S _2\right)=\frac{1}{2}$

First shelve has 5 physics and 3 biology books, i.e., total 8 books.

$\therefore P \left( P / S _1\right)=\frac{{ }^5 C_1}{{ }^8 C_1}=\frac{5}{8}$

Similarly, $P \left( P / S _2\right)=\frac{{ }^4 C_1}{{ }^6 C_1}=\frac{4}{6}=\frac{2}{3}$

$\therefore P(P)=P\left(S_1\right) \cdot P\left(P / S_1\right)+P\left(S_2\right) \cdot P\left(P / S_2\right)$

$\begin{aligned} & =\frac{1}{2} \times \frac{5}{8}+\frac{1}{2} \times \frac{2}{3} \\ & =\frac{31}{48}\end{aligned}$

17 white + 13 black = 30 shirts

4 white and 5 black are ‘PARTY WEAR’

A: Choosing a black shirt

$\therefore P(A)=\frac{{ }^{13} C_1}{30 C_1}=\frac{13}{30}$

B: Choosing a ‘PARTY WEAR’ shirt.

$\therefore P(B)=\frac{{ }^9 C _1}{{ }^{30} C _1}=\frac{9}{30}$

There are 5 black ‘PARTY WEAR’ shirts.

$\therefore P(A \cap B)=\frac{{ }^5 C _1}{3_0 C _1}=\frac{5}{30}$

∴ Required probability

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

$\begin{aligned} & =\frac{13}{30}+\frac{9}{30}-\frac{5}{30} \\ & =\frac{17}{30}\end{aligned}$

Two dice are thrown.

∴ n(S) = 36

Getting two numbers whose product is even, i.e., one of the two numbers must be even.

Let event A: Getting even number on first dice,

event B: Getting even number on second dice.

n(A) = 18, n(B) = 18, n(A ∩ B) = 9

Required probability = P(A ∩ B)

$\begin{aligned} & =\frac{n(A)+n(B)-n(A \cap B)}{n(S)} \\ & =\frac{18+18-9}{36} \\ & =\frac{3}{4}\end{aligned}$