11
Q:

Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?

A) 5/7 B) 1/5
C) 2/7 D) 2/35

Answer:   C) 2/7

Explanation:

P( only one of them will be selected)

             = p[(E and not F) or (F and not E)]

             = 

             = 

             =  

Q:

A letter iws takenout at random from 'ASSISTANT'  and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

A) 35/96 B) 19/90
C) 19/96 D) None of these
 
Answer & Explanation Answer: B) 19/90

Explanation:

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A =   = 1/45

Probability of choosing I =  = 1/45

Probability of choosing S =  = 1/10

Probability of choosing T =  = 1/15

Hence, Required probability = 

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45 1021
Q:

8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :

A) 8/39 B) 15/39
C) 12/13 D) None of these
 
Answer & Explanation Answer: B) 15/39

Explanation:

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize)

 = 

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36 1535
Q:

A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

A) 23/42 B) 19/42
C) 7/32 D) 16/39
 
Answer & Explanation Answer: B) 19/42

Explanation:

A red ball can be drawn in two mutually exclusive ways

(i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now,  = Probability of drawing a red ball when the first bag has been selected = 4/7

         = Probability of drawing a red ball when the second bag has been selected = 2/6

Using the law of total probability, we have 

P(red ball) = P(A) = 

                          = 

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9 1142
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 & Explanation Answer: C) 0.7

Explanation:

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

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32 922
Q:

A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

A) 1/4 B) 1/2
C) 3/4 D) 7/12
 
Answer & Explanation Answer: C) 3/4

Explanation:

Let A, B, C be the respective events of solving the problem and  be the respective events of not solving the problem. Then A, B, C are independent events

 are independent events

Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

 P( none  solves the problem) = P(not A) and (not B) and (not C)

                  = 

                  =                                                =  

                  = 

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

                = 

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21 7378