Advance Math Model Questions Set 2 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 2 EXERCISES
The following question based on Advance Math topic of quantitative aptitude
(a) dependent
(b) not pairwise independent
(c) independent
(d) none of the above.
The correct answers to the above question in:
Answer: (c)
Since, A and B are independent events.
∴ P(A ∩ B) = P(A)P(B)
Further since, A ∩ C, B ∩ C, A ∩ B ∩ C are subsets of C, we have
P(A ∩ C) ≤ P(C) = 0
P(B ∩ C) ≤ P(C) = 0
and P(A ∩ B ∩ C) ≤ P(C) = 0
⇒ P(A ∩ C) = 0 = P(A)P(C)
P(B ∩ C) = 0 = P(B)P(C)
P(A ∩ B ∩ C) = 0 = P(A)P(B)P(C).
Clearly A, B, C are pairwise independent as well as mutually independent. Thus, A,B,C are independent events.
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Read more model questions set 2 Based Quantitative Aptitude Questions and Answers
Question : 1
There are 6 different letters and 6 correspondingly addressed envelopes. If the letters are randomly put in the envelopes, what is the probability that exactly 5 letters go into the correctly addressed envelopes ?
a) $1/6$
b) $1/2$
c) Zero
d) $5/6$
Answer »Answer: (c)
As there are 6 letters and envelopes, so if exactly 5 are into correctly addressed envelopes, then the remaining 1 will automatically be placed in the correctly addressed envelope. Thus, the probability that exactly 5 go into the correctly addressed envelope is zero.
Question : 2
One bag contains 4 white balls and 2 black balls. Another bag contains 3 white balls and 5 black balls. If one ball is drawn from each bag, determine the probability that one ball is white and another is black.
a) $5/24$
b) $7/24$
c) $6/24$
d) $13/24$
Answer »Answer: (d)
Probability that first ball is white and second black
= (4/6) × (5/8) = 5/12
Probability that first ball is black and second white
= (2/6) × (3/8) = 1/8
These are mutually exclusive events hence the required probability
P = $5/{12} + 1/8 = {13}/{24}$.
Question : 3
The probability that the 13th day of a randomly chosen month is a Friday, is
a) $1/7$
b) $1/{84}$
c) $1/{12}$
d) $1/{13}$
Answer »Answer: (b)
Probability of selecting a month = $1/{12}$.
$13^{th}^$ day of the month is friday if its first day is sunday and the probability of this = $1/7$.
∴ Required probability = $1/{12}.1/7 = 1/{84}$.
Question : 4
A Chartered Accountant applies for a job in two firms X and Y. The probability of his being selected in firm X is 0.7 and being rejected at Y is 0.5 and the probability of atleast one of his applications being rejected is 0.6. What is the probability that he will be selected in one of the firms ?
a) 0.8
b) 0.4
c) 0.2
d) 0.7
Answer »Answer: (a)
P(x) = 0.7
= The probability of selected in firm X
P(y) = 1 – 0.5 = 0.5
= probability of selected in firms Y
P(X ∪ Y) = 1 – 0.6 = 0.4
Probability that he selected in one of the firms
= 0.7 + 0.5 – 0.4 = 0.8
Question : 5
A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behind. The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged ?
a) 6! × 1440
b) 18! × 2! × 1440
c) 18! × 1440
d) None of these
Answer »Answer: (c)
Two tallest boys can be arranged in 2! ways. Rest 18 can be arranged in 18! ways.
Girls can be arranged in 6! ways.
Total number of ways of arrangement = 2! × 18! × 6!
= 18! × 2 × 720 = 18! × 1440
Question : 6
Each of the 3 persons is to be given some identical items such that product of the numbers of items received by each of the three persons is equal to 30. In how many maximum different ways can this distribution be done ?
a) 24
b) 27
c) 21
d) 33
Answer »Answer: (b)
Suppose three people have been given a, b and c number of items.
Then, a × b × c = 30
Now, There can be 5 cases :
Case I : When one of them is given 30 items and rest two 1 item each.
So, number of ways for (30 × 1 × 1) = ${3!}/{2!}$ = 3
(As two of them have same number of items)
Case II : Similarly, number of ways for (10 × 3 × 1) = 3! = 6
Case III : Number of ways for (15 × 2 × 1) = 3! = 6
Case IV : Number of ways for (6 × 5 × 1) = 3! = 6
Case V : Number of ways for (5 × 3 × 2) = 3! = 6
Here, either of these 5 cases are possible.
Hence, total number of ways = 3 + 6 + 6 + 6 + 6 = 27
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