Aptitude ProbabilityPage 2
9. | | In a bag, there are 8 blue, 7 red and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green? | | (a)2/3 | | (b)3/4 | | (c)7/19 | | (d)8/21 |
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Answer is: DTotal number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green = event that the ball drawn is blue.
∴ n(E) = 8
∴ P(E) = 8/21.
10. | | Two dice are rolled simultaneously. Find the probability that one of them shows a number which is at least 4 and the other shows a number which is at most? | | (a)1/3 | | (b)1/2 | | (c)1/4 | | (d)1/5 |
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Answer is: BWhen two dice are thrown simultaneously,
total number of outcomes = 6 x 6 = 36.
Out of these, favorable cases are
(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3) i.e. 9.
∴ Required probability = 9 x (1/6) x (1/6) x 2 = 1/2.
11. | | Two dice are rolled simultaneously. Find the probability that the sum of the numbers on them is not 8? | | (a)31/36 | | (b)33/36 | | (c)35/36 | | (d)37/36 |
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Answer is: AProbability of sum of the numbers is not 8 = 1 – probability of sum of the numbers on them is 8.
favorable cases for sum of the number is 8 are (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
Probability (sum of the numbers on them is 8) = 5 (1/6) (1/6) = 5/36
Required probability = 1 – 5/36 = 31/36.
12. | | Five fair coins are tossed together. Find the probability of getting exactly 4 heads? | | (a)7/32 | | (b)9/32 | | (c)11/32 | | (d)5/32 |
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Answer is: DThe event of getting exactly 4 heads will be the combination of 4 heads and 1 tail.
Number of arrangements possible with this combination.
⇒ 5C4 = 5 !/4 ! 1 ! = 5
∴ Required probability = 5/25 = 5/32.
13. | | Five fair coins are tossed together. Find the probability of getting at most 3 heads? | | (a)13/16 | | (b)13/15 | | (c)15/13 | | (d)16/13 |
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Answer is: AProbability of getting at most 3 heads = 1 – probability of getting at least 4 heads
= 1 – [(probability of getting 4 heads) + (probability of getting 5 heads)]
= 1 – [5C4 (1/32) + 1/32]
= 1 – 6/32 = 13/16.
14. | | Two balls are drawn in succession without replacement from a box containing 3 black balls and 6 blue balls. Find the probability that the first is black and the second ball is blue? | | (a)1/6 | | (b)1/4 | | (c)1/8 | | (d)1/2 |
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Answer is: BThe first ball can be drawn in 9C1 =9 ways and as it is not replaced, the second ball can be drawn in 8C1 = 8 ways.
A black ball can be drawn in 3C1 = 3 ways in the first draw.
A blue ball can be drawn in 6C1 = 6 ways in the second draw.
∴ The required probability = (3/9) (6/8) = 1/4.
15. | | Two balls are drawn in succession without replacement from a box containing 3 black balls and 6 blue balls. Find the probability that both are blue? | | (a)7/12 | | (b)13/12 | | (c)5/12 | | (d)15/12 |
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Answer is: CThe first ball can be drawn in 9C1 =9 ways and as it is not replaced, the second ball can be drawn in 8C1 = 8 ways.
A blue ball can be drawn in 6C1 = 6 ways in the first draw.
Since this ball is not replaced, a blue ball can be drawn in 5C1 = 5 ways in the second draw.
∴ The required probability = (6/9) (5/8) = 5/12.
16. | | A bag contains 3 blue balls, 4 green balls and 5 red balls. A ball is drawn at random. Find the probability that it is not a green ball? | | (a)1/2 | | (b)2/3 | | (c)4/3 | | (d)5/3 |
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Answer is: BA ball can be drawn from 12 balls in 12C1 = 12 ways.
A ball other than a green ball (3 + 5 = 8) can be drawn in 8C1 = ways.
Hence, the probability of not drawing a green ball = 8C1/12C1 = 2/3.
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