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Probability

by gmattesttaker2 » Sat Sep 08, 2012 2:04 am
Hello,

This is from MGMAT Guide 5. P. 65. I was not clear with what the question was asking. Can you please assist? Thanks a lot - Sri

In a diving competition, each diver has a 20% chance of a perfect dive. The first perfect dive of the competition, but no subsequent dives, will receive a perfect score. If Janet is the third diver to dive, what is her chance of receiving a perfect score? (Assume that each diver can perform only one dive per turn)

Answer:
[spoiler](4/5) x (4/5) x (1/5) = 16/125[/spoiler]
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by Brent@GMATPrepNow » Sat Sep 08, 2012 6:44 am
gmattesttaker2 wrote:Hello,
In a diving competition, each diver has a 20% chance of a perfect dive. The first perfect dive of the competition, but no subsequent dives, will receive a perfect score. If Janet is the third diver to dive, what is her chance of receiving a perfect score? (Assume that each diver can perform only one dive per turn)

Answer:
[spoiler](4/5) x (4/5) x (1/5) = 16/125[/spoiler]
Here, we want P(Janet receives a perfect score)

At this point, we should always ask, "What needs to occur in order for this event to happen?"

Well, for Janet to receive a perfect score, she must be the first person in the competition to perform a perfect dive. Since she dives third, the first person must not dive perfectly AND the second person must not dive perfectly AND Janet must dive perfectly.

So, we can now write:
P(Janet receives a perfect score) = P(1st person does not dive perfectly AND 2nd person does not dive perfectly AND Janet dives perfectly)

We can now apply the AND probability rule and rewrite the probability as:
P(Janet receives a perfect score) = P(1st person does not dive perfectly) X P(2nd person does not dive perfectly) X P(Janet dives perfectly)

Now we're already told that the probability of a perfect dive is 20%, which is the same as 0.2 or 1/5.
So, the probability of a non-perfect dive is 80%, which is the same as 0.8 or 4/5.

So, we can write: P(Janet receives a perfect score) = (4/5)X(4/5)X(1/5)

Cheers,
Brent
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by gmattesttaker2 » Sun Sep 09, 2012 12:08 am
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote:Hello,
In a diving competition, each diver has a 20% chance of a perfect dive. The first perfect dive of the competition, but no subsequent dives, will receive a perfect score. If Janet is the third diver to dive, what is her chance of receiving a perfect score? (Assume that each diver can perform only one dive per turn)

Answer:
[spoiler](4/5) x (4/5) x (1/5) = 16/125[/spoiler]
Here, we want P(Janet receives a perfect score)

At this point, we should always ask, "What needs to occur in order for this event to happen?"

Well, for Janet to receive a perfect score, she must be the first person in the competition to perform a perfect dive. Since she dives third, the first person must not dive perfectly AND the second person must not dive perfectly AND Janet must dive perfectly.

So, we can now write:
P(Janet receives a perfect score) = P(1st person does not dive perfectly AND 2nd person does not dive perfectly AND Janet dives perfectly)

We can now apply the AND probability rule and rewrite the probability as:
P(Janet receives a perfect score) = P(1st person does not dive perfectly) X P(2nd person does not dive perfectly) X P(Janet dives perfectly)

Now we're already told that the probability of a perfect dive is 20%, which is the same as 0.2 or 1/5.
So, the probability of a non-perfect dive is 80%, which is the same as 0.8 or 4/5.

So, we can write: P(Janet receives a perfect score) = (4/5)X(4/5)X(1/5)

Cheers,
Brent
Hello Brent,

Thank you very much for your excellent explanation. As always, your explanation is very clear and easy to understand. Thanks again.

Best Regards,
Sri
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