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Probability Q

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Probability Q [#permalink] New post 16 Apr 2005, 19:53
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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Re: Probability Q [#permalink] New post 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).

Please post your solutions too.

Thanks.
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 [#permalink] New post 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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 [#permalink] New post 17 Apr 2005, 08:57
Yes the answers are correct!

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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 [#permalink] New post 17 Apr 2005, 09:01
1. A
2. B as 39/48 = 81%
  [#permalink] 17 Apr 2005, 09:01
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