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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

Answer & Explanation Answer: C) 0.7

Explanation:

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7

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Q:

A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

A) 1/4	B) 1/2
C) 3/4	D) 7/12

Answer & Explanation Answer: C) 3/4

Explanation:

Let A, B, C be the respective events of solving the problem and $A, B, C$ be the respective events of not solving the problem. Then A, B, C are independent event

$∴ A, B, C$ are independent events

Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

$P (A) = \frac{1}{2}, P (B) = \frac{2}{3}, P (C) = \frac{3}{4}$

$∴$ P( none solves the problem) = P(not A) and (not B) and (not C)

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

= $P (A) P (B) P (C)$ $[∵ A, B, C a r e I n d e p e n d e n t]$

= $\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}$

= $\frac{1}{4}$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $1 - \frac{1}{4}$ = 3/4

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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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