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 : 11
A bag has 13 red, 14 green and 15 white balls, $p_1$ is the probability of drawing exactly 2 white balls when four balls are drawn. Then the number of balls of each colour are doubled. Let $p_2$ be the probability of drawing 4 white balls when 8 ball are drawn, then
a) $p_1 > p_2$
b) $p_1 < p_2$
c) $p_1 = p_2$
d) None
Answer »Answer: (a)
$p_1 = {^{15}C_2}/{^{42}C_4} = {15 × 14 × 4!}/{2! × 42 × 41 × 40 × 39} = 1/{41 × 26}$ and
$p_2 = {^{30}C_4}/{^{84}C_8} = {15 × 14 × 13 × 12 × 8!}/{4! × 84 × 83 × 82 × ...... × 77}$
= ${15 × 14 × 13 × 12 × 8 × 7 × 6 × 5}/{84 × 83 × 82 × 81 × 79 × 78 × 77} < p_1$
⇒$p_1 > p_2$.
Question : 12
From a pack of 52 cards, two cards are drawn, the first being replaced before the second is drawn. What is the probability that the first is a diamond and the second is a king ?
a) $4/{13}$
b) $1/{52}$
c) $1/4$
d) $4/{15}$
Answer »Answer: (b)
Probability of getting a diamond, P(D) = ${13}/{52} = 1/4$
and probability to king, P(K) = $4/{52} = 1/{13}$
So, required probability = P(D).P(K)
= $1/4 × 1/{13} = 1/{52}$
Question : 13
In a given race the odds in favour of three horses A, B, C are 1 : 3; 1 : 4; 1 : 5 respectively. Assuming that dead head is impossible the probability that one of them wins is
a) ${37}/{60}$
b) $1/5$
c) $7/{60}$
d) $1/8$
Answer »Answer: (a)
Suppose $E_1 , E_2$ and $E_3$ are the events of winning the race by the horses A, B and C respectively
∴ $P(E_1) = 1/{1 + 3} = 1/4, P(E_2) = 1/{1 + 4} = 1/5$
$P(E_3) = 1/{1 + 5} = 1/6$
∴ Probability of winning the race by one of the horses A, B and C
= $P (E_1 or E_2 or E_3) = P(E_1) + P(E_2) + P(E_3)$
= $1/4 + 1/5 + 1/6 = {37}/{60}$
Question : 14
How many different letter arrangements can be made from the letter of the word EXTRA in such a way that the vowels are always together ?
a) 60
b) 40
c) 48
d) 30
Answer »Answer: (c)
Considering the two vowels E and A as one letter, the total no. of letters in the word 'EXTRA' is 4 which can be arranged in $^4P_4$ , i.e. 4! ways and the two vowels can be arranged among themselves in 2! ways.
∴ reqd. no. = 4! × 2! = 4 × 3 × 2 × 1 × 2 × 1 = 48
Question : 15
A,B,C and D are four towns any three of which are noncollinear. Then the number of ways to construct three roads each joining a pair of towns so that the roads do not form a triangle is
a) 8
b) 9
c) 7
d) More than 9
Answer »Answer: (d)
To construct 2 roads, three towns can be selected out of 4 in 4 ×3×2 = 24 ways. Now if the third road goes from the third town to the first town, a triangle is formed, and if it goes to the fourth town, a triangle is not formed. So there are 24 ways to form a triangle and 24 ways of avoiding a triangle.
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