5
Q:

# Two cards are drawn at random from a well - shuffled pack of 52 cards. what is the probability that either both are red or both are queens?

 A) 17/112 B) 55/221 C) 55/121 D) 33/221

Explanation:

n(S) = $\inline ^{52}\textrm{C}_{2}$ = 1326

Let  A = event of getting both red cards

and B = event of getting both queens

then $\inline A\cap B$ = event of getting two red queens

n(A) = $\inline ^{26}\textrm{C}_{2}$ = 325,   n(B) = $\inline ^{4}\textrm{C}_{2}$ = 6

$\inline n(A\cap B)= ^{2}\textrm{C}_{2}=1$

$\inline \therefore$ $\inline P(A)=\frac{325}{1326},\; \; P(B)=\frac{6}{1326}=\frac{1}{221}$

$\inline P(A\cap B)=\frac{1}{1326}$

$\inline \therefore$ P ( both red or both queens) = $\inline P(A\cup B)$

$\inline P(A)+P(B)-P(A\cap B)$

$\inline \frac{325}{1326}+\frac{1}{221}-\frac{1}{1326}$ = $\inline \frac{55}{221}$

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
Hence the answer is 1/14.

8 398
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

2 261
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.

1 208
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

10 650
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