2
Q:

# Tickets are numbered from 1 to 18 are mixed up together and then 9 tickets are drawn at random. Find the probability that the ticket has a number, which is a multiple of 2 or 3.

 A) 1/3 B) 3/5 C) 2/3 D) 5/6

Explanation:

S = { 1, 2, 3, 4, .....18 }

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

$\inline E3=(E1\cap E2) = \left \{ 6,12,18 \right \}$

=> n(E3) = 3

$\inline \therefore E = E1 \cup E2=E1+E2-E3$

=> n(E) = 9 + 6 - 3 =12

where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

$\inline \therefore P(E)= \frac{n(E)}{n(S)}= \frac{12}{18}=\frac{2}{3}$

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

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.

0 17
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

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'.

3 148
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]

Explanation:

Probability of atleast one failure
= 1 - no failure $\inline \fn_jvn \small \geq$ 31/32
= 1 - $\fn_jvn \small p^{5}$ $\inline \fn_jvn \small \geq$ 31/32
= $\fn_jvn \small p^{5}$ $\inline \fn_jvn \small \leq$ 1/32
= p $\inline \fn_jvn \small \leq$ 1/2
Also p $\inline \fn_jvn \small \geq$ 0
Hence p lies in [0,1/2].

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

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.

5 207
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

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