Probability Q : Quant Question Archive [LOCKED]
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Probability Q

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16 Apr 2005, 19:53
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hey Guys,

Know the Q's below are probability based .. But no clue how to go about solving them. If anyone could explain, that would be great! Thanks.

A nationwide survey showed that 65% of all children in the United States dislike eating vegetables. If 4 children are chosen at random, what is the probability that all 4 dislike eating vegetables? (Round your answer to the nearest percent)

a. 18%
b. 260%
c. 2%
d. none of the above

In New York state, 48% of all teenagers own a skateboard and 39% of all teenagers own a skateboard and roller blades. What is the probablity that a teenager owns roller blades given that the teenager owns a skateboard?

a. 87%
b. 81%
c. 123%
d. none of the above
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16 Apr 2005, 22:40
gix wrote:
Hey Guys,

Know the Q's below are probability based .. But no clue how to go about solving them. If anyone could explain, that would be great! Thanks.

A nationwide survey showed that 65% of all children in the United States dislike eating vegetables. If 4 children are chosen at random, what is the probability that all 4 dislike eating vegetables? (Round your answer to the nearest percent)

a. 18%
b. 260%
c. 2%
d. none of the above

In New York state, 48% of all teenagers own a skateboard and 39% of all teenagers own a skateboard and roller blades. What is the probablity that a teenager owns roller blades given that the teenager owns a skateboard?

a. 87%
b. 81%
c. 123%
d. none of the above

(1) The probability of a kid's disliking eating vegetables = 13/20.
The probability of 4 kids out of 4 disliking vegetables = (13/20)^4
= 0.178 = 18%
(D).

(2) Out of a sample 100 teeangers, 48 own skateboards, and of them 39 own roller blades as well. Thus for a given teenager having a skateboard, the probability of having roller blades = 39/48 = 13/16 = 0.712 = 71%
(D).

Thanks.
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17 Apr 2005, 01:44
I reached the same solutions for these problems.

(1) 65/100*65/100*65/100*65/100

(2) P(A|B)=P(A)/P(B) (when they are indipendent events)
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17 Apr 2005, 08:57

For the second one, your math actually works out to 81% and not 71% making the correct answer B.

Thanks very much!
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17 Apr 2005, 09:01
1. A
2. B as 39/48 = 81%
17 Apr 2005, 09:01
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