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Q:

Two cards are drawn at random from a well - shuffled pack of 52 cards. what is the probability that either both are red or both are queens?

A) 17/112	B) 55/221
C) 55/121	D) 33/221

Answer & Explanation Answer: B) 55/221

Explanation:

n(S) = ${}^{52}C_{2}$ = 1326

Let A = event of getting both red cards

and B = event of getting both queens

then $(A \cap B)$ = event of getting two red queens

n(A) = ${}^{26}C_{2}$ = 325, n(B) = ${}^{4}C_{2}$ = 6

$n (A \cap B) = {}^{2}C_{2} = 1$

$∴ P (A) = \frac{325}{1326}, P (B) = \frac{6}{1326}$

$P (A \cap B) = \frac{1}{1326}$

$∴$ P ( both red or both queens) = $P (A \cup B)$

= $P (A) + P (B) - P (A \cap B) =$ $\frac{325}{1326} + \frac{1}{221} - \frac{1}{1326}$ = $\frac{55}{221}$

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Q:

In a single throw of two dice , find the probability that neither a doublet nor a total of 8 will appear.

A) 7/15	B) 5/18
C) 13/18	D) 3/16

Answer & Explanation Answer: B) 5/18

Explanation:

n(S) = 36

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

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

$n (A) = 6, n (B) = 5, n (A \cap B) = 1$

$∴$ Required probability = $P (A \cup B)$

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

= $\frac{6}{36} + \frac{5}{36} - \frac{1}{36}$ = $\frac{5}{18}$

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Q:

The probability of occurance of two two events A and B are 1/4 and 1/2 respectively. The probability of their simultaneous occurrance is 7/50. Find the probability that neither A nor B occurs.

A) 25/99	B) 39/100
C) 61/100	D) 17/100

Answer & Explanation Answer: B) 39/100

Explanation:

P ( neither A nor B) = $P (A a n d B)$

= $P (A \cap B)$ = $P (A \cup B)$ = $1 - P (A \cup B)$

= $1 - \frac{61}{100} = \frac{39}{100}$

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Q:

A box contains 5 detective and 15 non-detective bulbs. Two bulbs are chosen at random. Find the probability that both the bulbs are non-defective.

A) 5/19	B) 3/20
C) 21/38	D) None of these

Answer & Explanation Answer: C) 21/38

Explanation:

n(S) = ${}^{20}C_{2}$ = 190

n(E) = ${}^{15}C_{2}$ = 105

Therefore, P(E) = 105/190 = 21/38

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Q:

A bag contains 8 red and 4 green balls. Find the probability that two balls are red and one ball is green when three balls are drawn at random.

A) 56/99	B) 112/495
C) 78/495	D) None of these

Answer & Explanation Answer: B) 112/495

Explanation:

$n (S) = {}^{12}C_{4} = 495$

$n (E) = {}^{8}C_{2} \times {}^{4}C_{1} = 112$

$∴ P (E) = \frac{112}{495}$

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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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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:

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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