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Q:

Tickets are numbered from 1 to 18 are mixed up together and then 9 tickets are drawn at random. Find the probability that the ticket has a number, which is a multiple of 2 or 3.

A) 1/3	B) 3/5
C) 2/3	D) 5/6

Answer & Explanation Answer: C) 2/3

Explanation:

S = { 1, 2, 3, 4, .....18 }

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

$E 3 = (E 1 \cap E 2) = {6, 12, 18}$

=> n(E3) = 3

$∴ E = E 1 \cup E 2 = E 1 + E 2 - E 3$

=> n(E) = 9 + 6 - 3 =12

where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

$∴ P (E) = \frac{n (E)}{n (S)} = \frac{12}{18} = \frac{2}{3}$

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Filed Under: Probability

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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Filed Under: Probability

Q:

One card is drawn from a pack of 52 cards , each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is neither a spade nor a king.

A) 0	B) 9/13
C) 1/2	D) 4/13

Answer & Explanation Answer: B) 9/13

Explanation:

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus n(E) = 13 + 3 = 16

Hence $n (\bar{E}) = 52 - 16 = 36$

Therefore, $P (E) = \frac{36}{52} = \frac{9}{13}$

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Filed Under: Probability

Q:

Two dice are rolled simultaneously. Find the probability of getting a total of 9.

A) 1/3	B) 1/9
C) 8/9	D) 9/10

Answer & Explanation Answer: B) 1/9

Explanation:

S = { (1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (2, 1), (2, 2),.........(6, 5), (6, 6) }

=> n(S) = 6 x 6 = 36

E = {(6, 3), (5, 4), (4, 5), (3, 6) }

=> n(E) = 4

Therefore, P(E) = 4/36 = 1/9

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Filed Under: Probability

Q:

An unbiased die is rolled. Find the probability of getting a multiple of 3?

A) 1/6	B) 1/3
C) 5/6	D) None of these

Answer & Explanation Answer: B) 1/3

Explanation:

S = { 1, 2, 3, 4, 5, 6 }

=> n(S) = 6

E = { 3, 6}

=> n(E) = 2

Therefore, P(E) = 2/6 =1/3

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Filed Under: Probability

Q:

Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails

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

Answer & Explanation Answer: D) 1/2

Explanation:

S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }

=> n(S) = 8

E = { HHH, HHT, HTH, THH }

=> n(E) = 4

P(E) = 4/8 = 1/2

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Q:

Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.

A) 28	B) 14
C) 21	D) 7

Answer & Explanation Answer: A) 28

Explanation:

Let x, y, z be the number of balls received by the three persons, then

$x \geq 5, y \geq 5, z \geq 5 a n d x + y + z = 21$

Let $u \geq 0, v \geq 0, w \geq 0 t h e n$

$∴$ x + y + z =21

$\Rightarrow$ u + 5 + v + 5 + w + 5 = 21

$\Rightarrow$ u + v + w = 6

$∴$ Total number of solutions = ${}^{6 + 3 - 1}C_{3 - 1} = {}^{8}C_{2} = 28$

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Filed Under: Permutations and Combinations

Q:

In How many ways is it possible to make a selection by taking any number of 15 fruits, namely 3 oranges, 5 apples and 7 mangoes?

A) 131	B) 191
C) 68	D) 3720

Answer & Explanation Answer: B) 191

Explanation:

Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind.

Hence, the required number of ways = [ (7+1) (5+1) (3+1) -1] =191

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Filed Under: Permutations and Combinations

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