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

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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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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Q:

The probabilities that A and B will tell the truth are 2 / 3 and 4 / 5 respectively . What is the probability that they agree with each other ?

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

Answer & Explanation Answer: A) 3/5

Explanation:

Let A be the event of A will tell truth. B be the event of B tell truth

$P (A) = \frac{2}{3}, P (A^{c}) = 1 - P (A) = \frac{1}{3}$

$P (B) = \frac{4}{5}, P (B^{c}) = 1 - P (B) = \frac{1}{5}$

When both agree then they say true or they say false together, that is

$A \cap B o r A^{c} \cap B^{c}$

Also these events will be mutually exclusive :

$P (A \cap B) + P (A^{c} \cap B^{c}) = P (A) P (B) P (C) = \frac{2}{3} * \frac{4}{5} + \frac{1}{3} * \frac{1}{5} = \frac{3}{5}$

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Q:

The probabilities that drivers A, B and C will drive home safely after consuming liquor are 2 / 5, 3 / 7 and 3 / 4, respectively. What is the probability that they will drive home safely after consuming liquor ?

A) 3/70	B) 4/70
C) 9/70	D) 1/50

Answer & Explanation Answer: C) 9/70

Explanation:

Let A be the event of driver A drive safely after consuming liquor.

Let B be the event of driver B drive safely after consuming liquor.

Let C be the event of driver C drive safely after consuming liquor.

P(A)=2/5,P(B)=3/7,P(C)=3/4

The events A, B and C are independent . Therefore,

$P (A \cap B \cap C) = P (A) P (B) P (C) = \frac{2}{5} * \frac{3}{7} * \frac{3}{4} = \frac{9}{70}$

Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9/70

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Q:

The odds favouring the event of a person hitting a target are 3 to 5. The odds against the event of another person hitting the target are 3 to 2. If each of them fire once at the target, find the probability that both of them hit the target.

A) 1/20	B) 4/20
C) 1/20	D) 3/20

Answer & Explanation Answer: D) 3/20

Explanation:

Let A be the event of first person hitting the target,

$P (A) = \frac{3}{3 + 5} = \frac{3}{8} (o d d i n f a v o u r)$

Let B be the event of Second person hitting a target.

$P (B) = \frac{2}{3 + 2} = \frac{2}{5} (o d d a g a i n s t)$

Since both events are independent and both will hit the target so,

$P (A \cap B) = P (A) P (B) = \frac{3}{8} \times \frac{2}{5} = \frac{3}{20}$

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Q:

A three-digit number is formed with the digits 1, 2, 3, 4, 5 at random. What is probability that number formed is divisible by 5 ?

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

Answer & Explanation Answer: B) 1/5

Explanation:

If a number ends with 5, 0 then the number will be divisible by 5
Here only 5 is present, end place will be fixed by 5 so,

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Q:

The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ?

A) 0.04	B) 0.043
C) 0.047	D) 0.05

Answer & Explanation Answer: B) 0.043

Explanation:

Vowels are A I A I O,

C A S T I G A T I O N

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

$n (E) = \frac{5 P_{5}}{2! * 2! (A, I a r e 2 t i m e s)} * \frac{6 P_{6}}{2! (T i s 2 t i m e s)} = 21600$

$n (S) = \frac{11!}{2! * 2! * 2! (A, I a r e 2 t i m e s)} = 4989600$

$\frac{21600}{4989600} = 0.043$

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Exam Prep: GRE

7 9723

Q:

A coin is tossed 5 times. What is the probability that head appears an odd number of times?

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

Answer & Explanation Answer: A) 1/2

Explanation:

The possible outcomes are as follows :

5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.

Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).

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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