12
Q:

# An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.

 A) 5/204 B) 1/204 C) 13/204 D) None of these

Explanation:

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then

Required probability = $\inline P\left ( A\cap B\cap C \right )$

$\inline P(A)P(\frac{B}{A})P(\frac{C}{A\cap B})$

Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

$\inline \therefore P(\frac{B}{A})=\frac{3}{17}$

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

$\inline \therefore P(\frac{C}{A\cap B})=\frac{2}{16}=\frac{1}{8}$

Hence the required probability = $\inline \frac{2}{9}\times \frac{3}{17}\times \frac{1}{8}=\frac{1}{204}$

Q:

The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?

 A) 1/7 B) 8! C) 7! D) 1/14

Explanation:

The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14

6 216
Q:

A fair six-sided die is rolled twice. What is the probability of getting 4 on the first roll and not getting 6 on the second roll ?

 A) 1/36 B) 5/36 C) 1/12 D) 1/9

Explanation:

The two events mentioned are independent.

The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.

P(getting first 4) = 1/6

P(no second 6) = 5/6

Therefore P(getting first 4 and no second 6) = 1/6 x 5/6 = 5/36

2 122
Q:

Let K and L be events on the same sample space, with P (K) = 0.8 and P (B) = 0.6. Are these two events are disjoint ?

 A) TRUE B) FALSE

Explanation:

These two events cannot be disjoint because P(K) + P(L) > 1.

P(AꓴB) = P(A) + P(B) - P(AꓵB).

An event is disjoint if P(A ꓵ B) = 0. If K and L are disjoint P(K ꓴ L) = 0.8 + 0.6 = 1.4

And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

1 82
Q:

What is the probability that a leap year has 53 Saturdays and 52 Sundays ?

 A) 1/7 B) 2/7 C) 1/2 D) 3/2

Explanation:

A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Friday, Saturday}

Required Probability = 1/7

8 456
Q:

Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is  ?

 A) 16/19 B) 1 C) 3/2 D) 17/20