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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:   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) = 

                          = 

Q:

If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is  ?

A) 1 B) 1/2
C) 0 D) 3/5
 
Answer & Explanation Answer: B) 1/2

Explanation:

The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.

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0 11
Q:

There are 10 Letters and 10 correspondingly 10 different Address. If the letter are put into envelope randomly, then find the Probability that Exactly 9 letters will at the Correct Address ?

A) 1/10 B) 1/9
C) 1 D) 0
 
Answer & Explanation Answer: D) 0

Explanation:

As we know we have 10 letter and 10 different address and one more information given that exactly 9 letter will at the correct address....so the remaining one letter automatically reach to their correct address
P(E) = favorable outcomes /total outcomes
Here favorable outcomes are '0'.
So probability is '0'.

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3 136
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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2 126
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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5 206
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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