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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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Filed Under: Probability
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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:

An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.

A) 5/204	B) 1/204
C) 13/204	D) None of these

Answer & Explanation Answer: B) 1/204

Explanation:

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then

Required probability = $P (A \cap B \cap C)$

= $P (A) P (\frac{B}{A}) P (\frac{C}{A \cap B})$

Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

$∴ P (\frac{B}{A}) = \frac{3}{17}$

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

$∴ P (\frac{C}{A \cap B}) = \frac{2}{16} = \frac{1}{8}$

Hence the required probability = $\frac{2}{9} \times \frac{3}{17} \times \frac{1}{8} = \frac{1}{204}$

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Q:

A coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely. Find the probability that the die shows a number greater than 4, if it is known that the first throw of the coin results in a tail

A) 1/3	B) 2/3
C) 2/5	D) 4/15

Answer & Explanation Answer: A) 1/3

Explanation:

Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }

Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,

A = { T5, T6 }

B = { T1, T2, T3, T4, T5, T6 }

Therefore, Required probability = $P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} = \frac{\frac{2}{8}}{\frac{6}{8}} = \frac{1}{3}$

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Q:

A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

A) 136/345	B) 17/87
C) 316/435	D) 158/435

Answer & Explanation Answer: C) 316/435

Explanation:

$n (s) = {}^{30}C_{2}$

Let A be the event of getting two oranges and

B be the event of getting two non-defective fruits.

and $(A \cap B)$ be the event of getting two non-defective oranges

$∴ P (A) = \frac{{}^{20}C_{2}}{{}^{30}C_{2}}, P (B) = \frac{{}^{22}C_{2}}{{}^{30}C_{2}} a n d P (A \cap B) = \frac{{}^{15}C_{2}}{{}^{30}C_{2}}$

$∴ P (A \cup B) = P (A) + P (B) - P (A \cap B)$

= $\frac{{}^{20}C_{2}}{{}^{30}C_{2}} + \frac{{}^{22}C_{2}}{{}^{30}C_{2}} - \frac{{}^{15}C_{2}}{{}^{30}C_{2}} = \frac{316}{435}$

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Q:

Four dice are thrown simultaneously. Find the probability that two of them show the same face and remaining two show the different faces.

A) 4/9	B) 5/9
C) 11/18	D) 7/9

Answer & Explanation Answer: B) 5/9

Explanation:

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in ${}^{6}C_{1}$ , ways.

Now select two distinct number out of remaining 5 numbers which can be done in ${}^{5}C_{2}$ ways. Thus these 4 numbers can be arranged in 4!/2! ways.

So, the number of ways in which two dice show the same face and the remaining two show different faces is

${}^{6}C_{1} \times {}^{5}C_{2} \times \frac{4!}{2!} = 720$

=> n(E) = 720

$∴ P (E) = \frac{720}{6^{4}} = \frac{5}{9}$

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Q:

A Committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to be included atleast one lady?

A) 123	B) 113
C) 246	D) 945

Answer & Explanation Answer: C) 246

Explanation:

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking.

(i) 1 lady out of 4 and 4 gentlemen out of 6

(ii) 2 ladies out of 4 and 3 gentlemen out of 6

(iii) 3 ladies out of 4 and 2 gentlemen out of 6

(iv) 4 ladies out of 4 and 1 gentlemen out of 6

In case I the number of ways = $C_{1}^{4} \times C_{4}^{6}$ = 4 x 15 = 60

In case II the number of ways = $C_{2}^{4} \times C_{3}^{6}$ = 6 x 20 = 120

In case III the number of ways = $C_{3}^{4} \times C_{2}^{6}$ = 4 x 15 = 60

In case IV the number of ways = $C_{4}^{4} \times C_{1}^{6}$ = 1 x 6 = 6

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

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Filed Under: Permutations and Combinations
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Q:

LAP BUT CAR SON HID

If the positions of the first and the third alphabets of each of the words are interchanged, which of the following would form a meaningful word in the new arrangement?

A) HID	B) SON
C) Both LAP and BUT	D) Both CAR and LAP

Answer & Explanation Answer: C) Both LAP and BUT

Explanation:

When positions of first and the third alphabets of each of the words are interchanged, we have PAL TUB RAC NOS DIH

Clearly, PAL and TUB are the only meaningful words. These words are obtained from LAP and BUT respectively.

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Filed Under: Alphabet Test

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