Advance Math Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (a)

Only one of A and B can be alive in the following, mutually exclusive ways.

$E_1$ A will die and B will live

$E_2$ B will die and A will live

So, required probability = $P(E_1) + P (E_2)$

= p(1 - q) + q (1 - p) = p + q - 2pq.

Answer: (b)

The student can choose 4 questions from first 5 questions or he can also choose 5 questions from the first five questions.

∴ No. of choices available to the student

= $^5C_4 × ^8C_6 + ^5C_5 × ^8C_5$ = 196.

Answer: (c)

The first card can be one of the 4 colours, the second can be one of the three and the third can be one of the two. The required probability is therefore

4 × ${13}/{52} × 3 × {13}/{51} × 2 × {13}/{50} = {169}/{425}$.

Answer: (b)

Required probability

= Probability that ball from bag A is red and both the balls from bag B are black + Probability that ball from bag A is black and one black and one red balls are drawn from bag

= ${^4C_1}/{^9C_1} × {^7C_2}/{^{10}C_2} + {^5C_1}/{^9C_1} × {^3C_1 × ^7C_1}/{^{10}C_2}$

= $4/9 × 7/{15} + 5/9 × 7/{15} = 7/{15}$

Answer: (a)

Exhaustive no. of cases = $6^3$

10 can appear on three dice either as distinct number as following (1, 3, 6) ; (1, 4, 5); (2, 3, 5) and each can occur in 3! ways. Or 10 can appear on three dice as repeated digits as following (2, 2, 6), (2, 4, 4), (3, 3, 4) and each can occur in ${3!}/{2!}$ ways.

∴ No. of favourable cases = 3 × 3! + 3 × ${3!}/{2!}$ = 27

Hence, the required probability = ${27}/{6^3} = 1/8$