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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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Filed Under: Probability

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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Filed Under: Probability

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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Filed Under: Probability

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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Filed Under: Probability

Q:

Tarsands containing large molecular hydrocarbons like crude oil are found in limited quantities in several parts of the world but the most important known occurrence is in:

A) U.S.A	B) Canada
C) Australia	D) Russia

Answer & Explanation Answer: B) Canada

Explanation:

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Filed Under: World Geography

Q:

The instrument of instructions contained in the Government of India Act, 1935 has been incorporated in the Constitution of India in the year 1950 as ______

A) Fundamental Rights	B) Directive Principles of the State Policy
C) Fundamental Duities	D) Emergency Provisions

Answer & Explanation Answer: D) Emergency Provisions

Explanation:

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Filed Under: Indian Politics

Q:

From a container of wine, a thief has stolen 15 litres of wine and replaced it with same quantity of water. He again repeated the same process. Thus, in three attempts the ratio of wine and water became 343 : 169. The initial amount of wine in the container was:

A) 75 litres	B) 100 litres
C) 150 litres	D) 120 litres

Answer & Explanation Answer: D) 120 litres

Explanation:

$\frac{w i n e (l e f t)}{w i n e (a d d e d)} = \frac{343}{169}$

It means $\frac{w i n e (l e f t)}{w i n e (i n i t i a l a m o u n t)} = \frac{343}{512}$ (since 343 + 169 = 512)

Thus, $343 x = 512 x {(1 - \frac{15}{k})}^{3}$

$\frac{343}{512} = {(\frac{7}{8})}^{3} = {(1 - \frac{15}{k})}^{3}$

$(1 - \frac{15}{k}) = \frac{7}{8} = (1 - \frac{1}{8})$

Thus the initial amount of wine was 120 liters.

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Filed Under: Alligation or Mixture
Exam Prep: GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

Q:

From a tank of petrol, which contains 200 litres of petrol, the seller replaces each time with kerosene when he sells 40 litres of petrol(or its mixture). Everytime he sells out only 40 litres of petrol(pure or impure). After replacing the petrol with kerosen 4th time, the total amount of kerosene in the mixture is

A) 81.92L	B) 96L
C) 118.08L	D) None of these

Answer & Explanation Answer: C) 118.08L

Explanation:

The amount of petrol left after 4 operations

= $200 \times {(1 - \frac{40}{200})}^{4}$

= $200 \times {(\frac{4}{5})}^{4}$

= $200 \times (\frac{256}{625})$

= 81.92 litres

Hence the amount of kerosene = 200 - 81.92 = 118. 08 litres

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Filed Under: Alligation or Mixture
Exam Prep: GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

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