Advance Math Model Questions Set 2 Practice Questions Answers Test with Solutions & More Shortcuts
Advance Math PRACTICE TEST [2 - EXERCISES]
Advance Math Model Questions Set 1
Advance Math Model Questions Set 2
Question : 1
The probability that at least one of the events A and B occurs is 0.7 and they occur simultaneously with probability 0.2. Then P($\ov{A}$) + P($\ov{B}$) =
a) 0.6
b) 1.1
c) 1.8
d) 0.4
Answer »Answer: (b)
We have P (A ∪ B) = 0.7 and P (A ∩ B) = 0.2
Now, P(A ∪ B) = P(A) + P(B) - P (A ∩ B)
⇒P(A) + P(B) = 0.9⇒1-P($\ov{A}$) + 1 - P($\ov{B}$) = 0.9
⇒P($\ov{A}$) + P($\ov{B}$) = 1.1
Question : 2
The number of ways of choosing a committee of 2 women and 3 men from 5 women and 6 men, if Mr. A refuses to serve on the committee if Mr. B is a member and Mr. B can only serve, if Miss C is the member of the committee, is
a) 84
b) 124
c) 60
d) None of these
Answer »Answer: (b)
(i) Miss C is taken
(1) B included ⇒A excluded ⇒$^4C_1 . ^4C_2$ = 24
(2) B excluded ⇒$^4C_1 . ^5C_3$ = 40
(ii) Miss C is not taken
⇒B does not comes ; $^4C_2 . ^5C_3$ = 60⇒Total = 124
Question : 3
The probabilities of four cricketers A, B, C and D scoring more than 50 runs in a match are $1/2 , 1/3 , 1/4$ and $1/{10}$ . It is known that exactly two of the players scored more than 50 runs in a particular match. The probability that these players were A and B is
a) $5/6$
b) $1/6$
c) ${27}/{65}$
d) None of these
Answer »Answer: (c)
Let $E_1$ be the event that exactly two players scored more than 50 runs then P$(E_1) = 1/2 × 1/3 × 3/4 × 9/{10} + 1/2 × 2/3 × 1/4 × 9/{10} + 1/2 × 2/3 × 3/4 × 1/{10} + 1/2 × 1/3 × 1/4 × 9/{10} + 1/2 × 1/3 × 3/4 × 1/{10} + 1/2 × 2/3 × 1/4 × 1/{10} = {65}/{240}$
Let $E_2$ be the event that A and B scored more than 50 runs, then P$(E_1 ∩ E_2) = 1/2 × 1/3 × 3/4 × 9/{10} = {27}/{240}$
∴ Desired probability
= $P(E_2/E_1) = {P(E_1 ∩ E_2)}/{P(E_1)} = {27}/{65}$
Question : 4
Each of two women and three men is to occupy one chair out of eight chairs, each of which numbered from 1 to 8. First, women are to occupy any two chairs from those numbered 1 to 4; and then the three men would occupy any, three chairs out of the remaining six chairs. What is the maximum number of different ways in which this can be done ?
a) 132
b) 1440
c) 40
d) 3660
Answer »Answer: (b)
2 Women can occupy 2 chairs out of the first four chairs in $^4P_2$ ways. 3 men can be arranged in the remaining 6 chairs in $^6P_3$ ways.
Hence, total no. of ways = $^4P_2 × ^6P_3$ = 1440
Question : 5
A man and his wife appear for an interview for two posts. The probability of the husband's selection is $1/7$ and that of the wife's selection is $1/5$ . The probability that only one of them will be selected is
a) $4/{35}$
b) $6/{35}$
c) $6/7$
d) $2/7$
Answer »Answer: (d)
Probability that only husband is selected
= P(H) P($\ov{W}$) = $1/7(1 - {1/5}) = 1/7 × 4/5 = 4/{35}$
Probability that only wife is selected
= P($\ov{H}$) P(W) = $(1 - {1/7})(1/5) = 6/7 × 1/5 = 6/{35}$
∴ Probability that only one of them is selected
=$4/{35} + 6/{35} = {10}/{35} = 2/7$
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