A sequence a, ax, ax2, ......, | Probability Questions & Answers

Q:

A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:

A) 123/897	B) 23/67
C) 7/44	D) 12/45

Answer & Explanation Answer: C) 7/44

Explanation:

Here, n(E) = $7 C_{1} \times 5 C_{1} \times 5 C_{1}$

And, n(S) = $12 C_{1} * 11 C_{1} * 10 C_{1}$

P(S) = $\frac{7 * 6 * 5}{12 * 11 * 10}$ = 7/44

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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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Q:

From a pack of 52 cards, 3 cards are drawn. What is the probability that one is ace, one is queen and one is jack?

A) 19/5525	B) 16/5525
C) 17/5525	D) 7/5525

Answer & Explanation Answer: B) 16/5525

Explanation:

Required probability:

$\frac{4 C_{1} * 4 C_{1} * 4 C_{1}}{52 C_{3}} = \frac{4 * 4 * 4}{22100} = \frac{16}{5525}$

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27 26255

Q:

In a race, the odd favour of cars P,Q,R,S are 1:3, 1:4, 1:5 and 1:6 respectively. Find the probability that one of them wins the race.

A) 319/420	B) 27/111
C) 114/121	D) 231/420

Answer & Explanation Answer: A) 319/420

Explanation:

$P (P) = \frac{1}{4}, P (Q) = \frac{1}{5}, P (R) = \frac{1}{6}, P (S) = \frac{1}{7}$
All the events are mutually exclusive hence,

Required probability = P(P)+P(Q)+P(R)+P(S)

$\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}$ = $\frac{319}{420}$

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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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Q:

If x is chosen at random from the set {1,2,3,4} and y is to be chosen at random from the set {5,6,7}, what is the probability that xy will be even?

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

Answer & Explanation Answer: B) 2/3

Explanation:

S ={(1,5),(1,6),(1,7),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7)}
Total element n(S)=12

xy will be even when even x or y or both will be even.
Events of x, y being even is E.
E ={(1,6),(2,5),(2,6),(2,7),(3,6),(4,5),(4,6),(4,7)}
n(E) = 8

P(E)=n(E)n(S)=812 = 4/3

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Q:

There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?

A) 1/2	B) 3/4
C) 4/7	D) 3/8

Answer & Explanation Answer: D) 3/8

Explanation:

Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways.

Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways.

Required probability =24/64 = 3/8

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44 25518

Q:

A card is drawn from a pack of 52 cards. The card is drawn at random. What is the probability that it is neither a spade nor a Jack?

A) 9/13	B) 4/13
C) 10/13	D) 8/13

Answer & Explanation Answer: A) 9/13

Explanation:

There are 13 spade and 3 more jack

Probability of getting spade or a jack:
=13+352=1652=413

So probability of getting neither spade nor a jack:
=1−413 = 9/13

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A) axn2+1	B) axn2-1
C) axn-1	D) axn2

A sequence $a, ax, {ax}^{2}, . . . . . ., {ax}^{n}$ , has odd number of terms. Then the median is

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A sequence a, ax, ax2, ......, axn, has odd number of terms. Then the median is

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A sequence $a, ax, {ax}^{2}, . . . . . ., {ax}^{n}$ , has odd number of terms. Then the median is