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) $5/24$
(b) $7/24$
(c) $6/24$
(d) $13/24$
The correct answers to the above question in:
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 = $5/{12} + 1/8 = {13}/{24}$.
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Read more model questions set 2 Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?
a) 10
b) 9
c) 5
d) 15
Answer »Answer: (b)
For each person to know all the secrets the communication has to be between the Englishmen (who knows say E1 French) and one Frenchmen (say F1). The other two in each case will communicate with E1 & F1 respectively. So for minimum no. of calls, E2 gives information to E1 & receives it after E1 interacts with F1. So 2 calls for each of the four E2, E3, F2 and F3, i.e., 8 calls +1 call (between E1 & F1). Hence 9 calls in all.
Question : 3
A programmer noted the results of attempting to run 20 programs. The results showed that 2 programs ran correctly in the first attempt, 7 ran correctly in the second attempt, 5 ran correctly in the third attempt, 4 ran correctly in the fourth attempt and 2 ran correctly in the fifth attempt. What is the probability that his next programme will run correctly on the third run ?
a) $1/3$
b) $1/6$
c) $1/4$
d) $1/5$
Answer »Answer: (c)
Total number of attempts = 20
Favourable no. of attempts = 5
Required probability (running the program correctly in the third run) = $5/{20} = 1/4$
Question : 4
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 : 5
If A and B are two independent events and P(C) = 0, then A, B, C are :
a) dependent
b) not pairwise independent
c) independent
d) none of the above.
Answer »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.
Question : 6
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
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