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Q:

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

A) 564	B) 735
C) 756	D) 657

Answer & Explanation Answer: C) 756

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways= $(7 C_{3} * 6 C_{2}) + (7 C_{4} * 6 C_{1}) + 7 C_{5}$ = 756.

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Filed Under: Permutations and Combinations

Q:

In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, what is the probability that he. has offered English or Hindi ?

A) 1/2	B) 3/4
C) 4/5	D) 2/5

Answer & Explanation Answer: D) 2/5

Explanation:

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Filed Under: Probability

Q:

A man and his wife appear in an interview for two vacancies in the same post. The probability of husband's selection is (1/7) and the probability of wife's selection is (1/5). What is the probability that only one of them is selected ?

A) 2/7	B) 1/7
C) 3/4	D) 4/5

Answer & Explanation Answer: A) 2/7

Explanation:

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Filed Under: Probability

Q:

A number is selected at random from the numbers 1 to 30. What is the probability that it is divisible by either 3 or 7 ?

A) 1/30	B) 2/30
C) 1/25	D) 1/2

Answer & Explanation Answer: A) 1/30

Explanation:

Let A be event of selecting a number divisible by 3. B be the event of selecting a number divisible by 7.

A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 }, so n(A)=10
B = { 7, 14, 21, 28 }, n(B)= 4

Since A and B are not mutually exclusive So :

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Filed Under: Probability

Q:

A brother and a sister appear for an interview against two vacant posts in an office. The probability of the brother’s selection is 1/5 and that of the sister’s selection is 1/3. What is the probability that only one of them is selected?

A) 1/5	B) 3/4
C) 2/5	D) 3/5

Answer & Explanation Answer: C) 2/5

Explanation:

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) + (Prob. that brother is not selected) × (Prob. that sister is selected)

= $(\frac{1}{5} * \frac{2}{3}) + (\frac{4}{5} * \frac{1}{3})$ = $\frac{2}{5}$

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Filed Under: Probability

Q:

In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?

A) 0.25	B) 0.5
C) 0.75	D) 0.40

Answer & Explanation Answer: D) 0.40

Explanation:

250 numbers between 101 and 350 i.e. n(S)=250

n(E)=100th digits of 2 = 299−199 = 100

P(E)= n(E)/n(S) = 100/250 = 0.40

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Filed Under: Probability

Q:

Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.

A) 19/34	B) 5/4
C) 20/34	D) 25/34

Answer & Explanation Answer: D) 25/34

Explanation:

The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl.

So the required probability:
= $(\frac{8}{17} * \frac{9}{16}) + (\frac{9}{17} * \frac{8}{16}) + (\frac{8}{17} * \frac{7}{16})$

= $\frac{9}{34} + \frac{9}{34} + \frac{7}{34}$
= $\frac{25}{34}$

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Filed Under: Probability

Q:

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

A) 1/6	B) 1/3
C) 1/2	D) 1/4

Answer & Explanation Answer: C) 1/2

Explanation:

P(odd) = P (even) = $\frac{1}{2}$ 1(because there are 50 odd and 50 even numbers)

Sum or the three numbers can be odd only under the following 4 scenarios:

Odd + Odd + Odd = $\frac{1}{2} * \frac{1}{2} * \frac{1}{2}$ = $\frac{1}{8}$

Odd + Even + Even = $\frac{1}{2} * \frac{1}{2} * \frac{1}{2}$ = $\frac{1}{8}$

Even + Odd + Even = $\frac{1}{2} * \frac{1}{2} * \frac{1}{2}$ = $\frac{1}{8}$

Even + Even + Odd = $\frac{1}{2} * \frac{1}{2} * \frac{1}{2}$ = $\frac{1}{8}$

Other combinations of odd and even will give even numbers.

Adding up the 4 scenarios above:

= $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ = $\frac{4}{8}$ = $\frac{1}{2}$

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Filed Under: Probability

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