Advance Math Model Questions Set 1 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 : 11
Ram and Shyam appear for an interview for two vacancies in an organisation for the same post. The probabilities of their selection are 1/6 and 2/5 respectively. What is the probability that none of them will be selected ?
a) $1/5$
b) $1/2$
c) $5/6$
d) $3/5$
Answer »Answer: (b)
Required probability = $(1 - 1/6) × (1- 2/5) = 5/6 × 3/5 = 1/2$
Question : 12
A car is parked by an owner amongst 25 cars in a row, not at either end. On his return he finds that exactly 15 places are still occupied. The probability that both the neighbouring places are empty is
a) ${15}/{184}$
b) ${15}/{92}$
c) ${91}/{276}$
d) None
Answer »Answer: (b)
Exhaustive number of cases = $^{24}C_{14}$
Favourable cases = $^{22}C_{14}$
Probability = ${^{22}C_{14}}/{^{24}C_14} = {15}/{92}$
Question : 13
To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants, and 2 managers from among 4 applicants. What is the total number of ways in which she can make her selection ?
a) 132
b) 120
c) 1,490
d) 60
Answer »Answer: (b)
Required no. of the ways = $^6C_3 × ^4C_2$ = 20 × 6 = 120
Question : 14
A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective, is:
a) $7/{19}$
b) ${12}/{19}$
c) $4/{19}$
d) ${21}/{95}$
Answer »Answer: (a)
P (None is defective) = ${^{16}C_2}/{^{20}C_2} = ({16 × 15}/{2 × 1} × {2 × 1}/{20 × 19})$
= ${12}/{19}.$
P (at least one is defective) = 1 - ${12}/{19} = 7/{19}$
Question : 15
3 digits are chosen at random from 1,2,3,4,5,6,7,8 and 9 without repeating any digit. What is the probability that their product is odd?
a) $5/108$
b) $5/42$
c) $2/3$
d) $8/42$
Answer »Answer: (b)
Let E be the event of selecting the three numbers such that their product is odd and S be the sample space. For the product to be odd, 3 numbers choosen must be odd.
∴ n(E) = $^5C_3$
= n(S) = $^9C_3$
∴ P(E) = ${n(E)}/{n(S)} = {^5C_3}/{^9C_3} = 5/{42}$
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