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91
In hypergeometric distribution, the trials are
Discuss
Answer & Solution
Answer: Option A
Solution:
Dependent
92
For a random variable X, E(X) is
Discuss
Answer & Solution
Answer: Option C
Solution:
Arithmetic Mean (AM)
93
Probability distribution of a random variable is also known as
Discuss
Answer & Solution
Answer: Option B
Solution:
Probability Function
94
There are 20 balls in a bag which are numbered 1, 2, 3 . . . . . 20. Find the probability that the number marked on the ball taken out of the bag is divisible by 3 or 5.
Discuss
Answer & Solution
Answer: Option B
Solution:
Number divisible by 3 = 6
Number divisible by 5 = 4
Number divisible by 3 or 5 = 1
Total number = 6 + 4 - 1 = 9
Probability = $$\frac{9}{{20}}$$
95
What is the probability that a two digit number is not a prime number when a number is chosen at random?
Discuss
Answer & Solution
Answer: Option B
Solution:
Two digit number = 10 . . . . . 99 = 90 number
Prime number between 1 to 100 = 25
Prime number between 1 to 10 = 2, 3, 5, 7 = 4
Prime number between 1 to 99 = 25 - 4 = 21
$${\text{Probability}} = \frac{{90 - 21}}{{90}} = \frac{{69}}{{90}} = \frac{{23}}{{30}}$$
96
A letter of English alphabet is chosen at random, Probability of getting a vowel is-
Discuss
Answer & Solution
Answer: Option A
Solution:
$${\text{P}}\left( {\text{V}} \right) = \frac{5}{{26}}$$
97
When a coined is tossed once, what are the probability of coming Head?
Discuss
Answer & Solution
Answer: Option B
Solution:
$${\text{P}} = \frac{1}{2}$$
98
What will be the probability to remove face card from card deck?
Discuss
Answer & Solution
Answer: Option C
Solution:
Total card = 52
Face card = 12
$${\text{P}} = \frac{{12}}{{52}} = \frac{3}{{13}}$$
99
A bag contains cards numbered between 33 and 92. If one card is drawn from the bag, the probability of the number of the drawn card is a perfect square is-
Discuss
Answer & Solution
Answer: Option D
Solution:
Perfect square between 33 to 92 = 36, 49, 64, 81
Total number = 92 - 33 = 59
$${\text{P}} = \frac{4}{{59}}$$
100
When a pair of dice is thrown, what is the probability of the sum of numbers being odd?
Discuss
Answer & Solution
Answer: Option D
Solution:
Total events = 62 = 36
Sum odd = 1, 3, 5, 7, 9, 11
3 = (1, 2)(2, 1)
5 = (1, 4)(4, 1)(2, 3)(3, 2)
7 = (1, 6)(6, 1)(2, 5)(5, 2)(3, 4)(4, 3)
9 = (3, 6)(6, 3)(4, 5)(5, 4)
11 = (6, 5)(5, 6)
$${\text{Probability}} = \frac{{18}}{{36}} = \frac{1}{2} = 0.5$$