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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:

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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