MHTCET 14 May 2023 Shift 1 Solved Paper

The probability that the first critic favours the book is P(A)=‌

2

2+5

=‌

2

7

∴‌‌P(A′)=1−‌

2

7

=‌

5

7

The probability that the second critic favours the book is P(B)=‌

3

3+4

=‌

3

7

∴‌‌P(B′)=1−‌

3

7

=‌

4

7

The probability that the third critic favours the book is P(C)=‌

4

4+3

=‌

4

7

∴‌‌P(C′)=1−‌

4

7

=‌

3

7

∴ Majority will be in favour of the book if at least two critics favour the book. Hence, the probability is
‌P(A∩B∩C′)+P(A∩B′∩C)
‌‌‌+P(A′∩B∩C)+P(A∩B∩C)
‌=P(A)⋅P(B)⋅P(C′)+P(A)⋅P(B′)⋅P(C)
‌‌‌+P(A′)⋅P(B)⋅P(C)+P(A)⋅P(B)⋅P(C)
‌=‌

18

343

+‌

32

343

+‌

60

343

+‌

24

343

‌=‌

134

343