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

Q:

Out of sixty students, there are 14 who are taking Economics and 29 who are taking Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class ?

A) 8/15	B) 7/15
C) 1/15	D) 4/15

Answer & Explanation Answer: B) 7/15

Explanation:

Given total students in the class = 60
Students who are taking Economics = 24 and
Students who are taking Calculus = 32
Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4
Students who are taking calculus only = 32 - 4 = 28
probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.

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Q:

Three unbiased coins are tossed. What is the probability of getting at most two heads ?

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

Answer & Explanation Answer: D) 3/4

Explanation:

Let S be the sample space.
Here n(S)= $2^{3}$ = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

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Q:

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5 ?

A) 2/3	B) 1/2
C) 7/8	D) 4/5

Answer & Explanation Answer: B) 1/2

Explanation:

Multiples of 3 below 20 are 3, 6, 9, 12, 15, 18
Multiples of 5 below 20 are 5, 10, 15, 20
Required number of possibilities = 10
Total number of possibilities = 20
Required probability = 10/20 = 1/2.

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Q:

A letter is takenout at random from 'ASSISTANT' and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

A) 35/96	B) 19/90
C) 19/96	D) None of these

Answer & Explanation Answer: B) 19/90

Explanation:

$A S S I S T A N T \to A A I N S S S T T$

$S T A T I S T I C S \to A C I I S S S T T T$

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A = $\frac{2_{C_{1}}}{9_{C_{1}}} \times \frac{1_{C_{1}}}{10_{C_{1}}}$ = 1/45

Probability of choosing I = $\frac{1}{9_{C_{1}}} \times \frac{2_{C_{1}}}{10_{C_{1}}}$ = 1/45

Probability of choosing S = $\frac{3_{C_{1}}}{9_{C_{1}}} \times \frac{3_{C_{1}}}{10_{C_{1}}}$ = 1/10

Probability of choosing T = $\frac{2_{C_{1}}}{9_{C_{1}}} \times \frac{3_{C_{1}}}{10_{C_{1}}}$ = 1/15

Hence, Required probability = $\frac{1}{45} + \frac{1}{45} + \frac{1}{10} + \frac{1}{15} = \frac{19}{90}$

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Q:

8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :

A) 8/39	B) 15/39
C) 12/13	D) None of these

Answer & Explanation Answer: B) 15/39

Explanation:

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize)

= $1 - (\frac{16_{C_{1}} \times 14_{C_{1}} \times 12_{C_{1}} \times 10_{C_{1}}}{16_{C_{4}}}) = \frac{15}{39}$

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Q:

A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

A) 23/42	B) 19/42
C) 7/32	D) 16/39

Answer & Explanation Answer: B) 19/42

Explanation:

A red ball can be drawn in two mutually exclusive ways

(i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore

P(E1) = 1/2 and P(E2) = 1/2

Now, $P (\frac{A}{E_{1}})$ = Probability of drawing a red ball when the first bag has been selected = 4/7

$P (\frac{A}{E_{2}})$ = Probability of drawing a red ball when the second bag has been selected = 2/6

Using the law of total probability, we have

P(red ball) = P(A) = $P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}})$

= $\frac{1}{2} \times \frac{4}{7} + \frac{1}{2} \times \frac{2}{6} = \frac{19}{42}$

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Q:

Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?

A) 5/7	B) 1/5
C) 2/7	D) 2/35

Answer & Explanation Answer: C) 2/7

Explanation:

P( only one of them will be selected) = p[(E and not F) or (F and not E)]

= $P [(E \cap F) \cup (F \cap E)]$

= $P (E) P (F) + P (F) P (E)$

= $\frac{1}{7} \times \frac{4}{5} + \frac{1}{5} \times \frac{6}{7} = \frac{2}{7}$

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Q:

The probabilities that a student will receive an A, B, C or D grade are 0.4, 0.3 , 0.2 and 0.1 respectively. Find the probability that a student will receive Atleast B grade.

A) 0.21	B) 0.3
C) 0.7	D) None of these

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