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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

Answer:   B) 1/204



Explanation:

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

Required probability = 

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

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.

Hence the required probability = 

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
 
Answer & Explanation Answer: B) 5/36

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

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1 45
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
Answer & Explanation Answer: 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.

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1 16
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
 
Answer & Explanation Answer: A) 1/7

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

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5 264
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
 
Answer & Explanation Answer: D) 17/20

Explanation:

n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
P(E U P) = 10/20 + 8/20 - 1/20 = 17/20

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3 216
Q:

The probability that a number selected at random from the first 100 natural numbers is a composite number is  ?

A) 3/2 B) 2/3
C) 1/2 D) 34/7
 
Answer & Explanation Answer: A) 3/2

Explanation:

The number of exhaustive events = 100 C₁ = 100.

We have 25 primes from 1 to 100.

Number of favourable cases are 75.

Required probability = 75/50 = 3/2.

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2 233