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Q:

The probabilities that a student will receive an A, B, C or D grade are 0.4, 0.3 , 0.2 and 0.1 respectively. Find the probability that a student will receive Atleast B grade.

A) 0.21 B) 0.3
C) 0.7 D) None of these

Answer:   C) 0.7

Explanation:

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7

Q:

Consider 5 independent Bernoulli’s trials each with probability of success p. If the probability of at least one failure is greater than or equal to 31/32 , then p lies in the interval ?

A) [1, 32] B) (0, 1)
C) [1, 1/2] D) (1, 1/2]
 
Answer & Explanation Answer: C) [1, 1/2]

Explanation:

Probability of atleast one failure
= 1 - no failure  31/32
= 1 -   31/32
=   1/32
= p  1/2
Also p  0
Hence p lies in [0,1/2].

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1 43
Q:

One lady has 2 children, one of her child is boy, what is the probability of having both are boys ?

A) 1/3 B) 1/2
C) 2/3 D) 2/5
 
Answer & Explanation Answer: A) 1/3

Explanation:

In a family with 2 children there are four possibilities:
1) the first child is a boy and the second child is a boy (bb)
2) the first child is a boy and the second child is a girl (bg)
3) the first child is a girl and the second child is a boy (gb)
4) the first child is a girl and the second child is a girl (gg)

But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb).
n(E)= both are boys=BB=1
n(S)= 3
Required probability P = n(E)/n(S) = 1/3.

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2 104
Q:

fifteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together ?

A) 3/91 B) 2/73
C) 1/91 D) 3/73
 
Answer & Explanation Answer: A) 3/91

Explanation:

In a circle of n different persons, the total number of arrangements possible = (n - 1)!
Total number of arrangements = n(S) = (15 – 1)! = 14 !
Taking three persons as a unit, total persons = 13 (in 4 units)
Therefore no. of ways for these 13 persons to around the circular table = (13 - 1)! = 12!
In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =
n(E) = 12! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!
= 12!x3x2 / 14x13x12! = 6/14x13 = 3/91

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5 127
Q:

The Manager of a company accepts only one employees leave request for a particular day. If five employees namely Roshan, Mahesh, Sripad, Laxmipriya and Shreyan applied for the leave on the occasion of Diwali. What is the probability that Laxmi priya’s leave request will be approved ?

A) 1 B) 1/5
C) 5 D) 4/5
 
Answer & Explanation Answer: B) 1/5

Explanation:

Number of applicants = 5
On a day, only 1 leave is approved.
Now favourable events =  1 of 5 applicants is approved
Probability that Laxmi priya's leave is granted = 1/5.

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5 103
Q:

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 15 ?

A) 6/19 B) 3/10
C) 7/10 D) 6/17
 
Answer & Explanation Answer: B) 3/10

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

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5 141