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

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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Q:

A six-sided die with faces numbered 1 through 6 is rolled three times. What is the probability that the face with the number 6 on it will not face upward on all the three rolls?

A) 215/216	B) 1/216
C) 215/256	D) 1/52

Answer & Explanation Answer: A) 215/216

Explanation:

Probability of getting 6 at the top once =1/6

Probability of getting 6 at the top three times =1/6 x 1/6 x 1/6 =216

Probability of no getting 6 at he top any time = 1 - 1/216 = 215/216

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Q:

I forgot the last digit of a 7-digit telephone number. If 1 randomly dial the final 3 digits after correctly dialing the first four, then what is the chance of dialing the correct number?

A) 1/999	B) 1/1001
C) 1/1000	D) 4/1000

Answer & Explanation Answer: C) 1/1000

Explanation:

It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways.
Hence required probability = ${(\frac{1}{10})}^{3}$ = 1/1000

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Q:

What is the probability of getting at least one six in a single throw of three unbiased dice?

A) 1/36	B) 91/256
C) 13/256	D) 43/256

Answer & Explanation Answer: B) 91/256

Explanation:

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5×5×5=125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

=216−125=91

The required probability = 91/256

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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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Q:

Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King

A) 1/52	B) 1/26
C) 1/13	D) 1/2

Answer & Explanation Answer: B) 1/26

Explanation:

Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)

= $[\frac{12 C_{1} * 1 C_{1}}{52 C_{2}}] + [\frac{13 C_{1} * 3 C_{1}}{52 C_{2}}]$

= $[12 * \frac{2}{52 * 51}] + [13 * 3 * \frac{2}{52 * 51}]$ = $\frac{24 + 78}{52 * 51}$ = $\frac{1}{26}$

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Q:

The probability of success of three students X, Y and Z in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.

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

Answer & Explanation Answer: C) 1/6

Explanation:

P(X) = $\frac{1}{5}$ , P(Y) = $\frac{1}{4}$ , P(Z) = $\frac{1}{3}$

Required probability:

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

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

= $\frac{4}{60} + \frac{3}{60} + \frac{2}{60} + \frac{1}{60}$ = $\frac{10}{60}$ = $\frac{1}{6}$

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Q:

You toss a coin AND roll a die. What is the probability of getting a tail and a 4 on the die?

A) 1/2	B) 1/12
C) 2/3	D) 3/4

Answer & Explanation Answer: B) 1/12

Explanation:

Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16

Required probability =(12)×(16)
= 1/12

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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