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# In a diving competition, each diver has a 20% chance of a perfect dive

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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
Hi, I understand why it's 16/125, but could someone tell me why it's not 32/125?

Since Janet is the third driver, if we assume that there are 3 drivers (other 2 being X and Y), the possible cases could be - J-X-Y and J-Y-X, where in the first case X comes 2nd and in the second case Y comes 2nd.

In this case probability would be

Event I: JXY - (1/5)(4/5)(4/5) = 16/125
Event II: JYX - (1/5)(4/5)(4/5) = 16/125

Probability = Event I + Event II = 16+16/125 = 32/125. Where am i going wrong?

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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
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All other nonperfect divers are treated as identical events

4/5×4/5×1/5
=16/125

D. 16/125
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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
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Bunuel wrote:
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.)

A. 1/5
B. 1/15
C. 4/25
D. 16/125
E. 61/125

Kudos for a correct solution.

Probability = Favourable Outcomes / Total Outcomes = 1 - (Unfavourable Outcomes / Total Outcomes)

Given: Probability of Perfect dive = 0.2 = 20/100 = 1/5
i.e. Probability of Non-Perfect dive = 0.8 = 80/100 = 4/5

The probability of Janet to dive and get a perfect score depends on that other other previous two dives must be Imperfect

Method-1:

i.e. Probability of First two being Imperfect and Third being Perfect dive = (4/5)*(4/5)*(1/5) = 16/125

Method-2:

i.e. Probability of all the three Dives being Imperfect dives = (4/5)*(4/5)*(4/5) = 64/125

and Probability of First dive being Imperfect and Second being Perfect dive = (4/5)*(1/5) = 4/25

and Probability of First dive being Perfect dive = (1/5) = 1/5

I.e. Total Unfavourable probability = (64/125)+(4/25)+(1/5) = (64+20+25)/125 = 109/125

i.e. Total Favorable probability = 1-(109/125) = 16/125

Please Note: Second method is only to develop understanding and the best method to solve this question is by using the first method

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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
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FireStorm wrote:
Hi, I understand why it's 16/125, but could someone tell me why it's not 32/125?

Since Janet is the third driver, if we assume that there are 3 drivers (other 2 being X and Y), the possible cases could be - J-X-Y and J-Y-X, where in the first case X comes 2nd and in the second case Y comes 2nd.

In this case probability would be

Event I: JXY - (1/5)(4/5)(4/5) = 16/125
Event II: JYX - (1/5)(4/5)(4/5) = 16/125

Probability = Event I + Event II = 16+16/125 = 32/125. Where am i going wrong?

You have taken the wrong set of events

Unfavourable Set of events are as follows

Event 1: First Dive is perfect = 1/5
Event 2: First Dive is Imperfect and second is Perfect = (4/5)*(1/5) = 4/25
Event 3: All three dives are Imperfect = (4/5)*(4/5)*(4/5) = 64/125

Forth and only possible remaining event (Favourable Event)
Event 4: First Two dives are Imperfect and Third is Perfect = (4/5)*(4/5)*(1/5) = 16/125

Favourable + Unfavourable Probability = (1/5)+(4/25)+(64/125)+(16/125) = 1

I have solved this as Method-2 in my above Comment as well.

I hope this helps!
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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
If janet wants to win , then first two drivers should not score perfect and janet should score perfect

Hence , 4/5 *4/5* 1/5 = 16/125

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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
FireStorm wrote:
Hi, I understand why it's 16/125, but could someone tell me why it's not 32/125?

Since Janet is the third driver, if we assume that there are 3 drivers (other 2 being X and Y), the possible cases could be - J-X-Y and J-Y-X, where in the first case X comes 2nd and in the second case Y comes 2nd.

In this case probability would be

Event I: JXY - (1/5)(4/5)(4/5) = 16/125
Event II: JYX - (1/5)(4/5)(4/5) = 16/125

Probability = Event I + Event II = 16+16/125 = 32/125. Where am i going wrong?

It can not be JXY or JYX as it given that Janet is the 3rd diver to dive. So both these cases are ruled out.
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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
Bunuel wrote:
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.)

A. 1/5
B. 1/15
C. 4/25
D. 16/125
E. 61/125

Kudos for a correct solution.

Probability of a perfect dive = P(D) = 0.2
Probability of a no perfect dive = 1-0.2 = 0.8

Thus, for Jane to get the perfect score, the first 2 divers should NOT get a perfect score (as it is given that only 1 diver can get a perfect score!)

Thus, the final probability is = 0.8*0.8*0.2 = 16/125, hence D is the correct answer.
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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
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Probability of a Perfect drive =1/5 .Janet is the third driver .So the first two drivers must have failed to get a perfect drive
Probability=(Probability Of first Failure )*(Probability Of Second Failure)*Probability of Success
Probability =(4/5)*(4/5)*(1/5)
Option D

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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
FireStorm wrote:
Hi, I understand why it's 16/125, but could someone tell me why it's not 32/125?

Since Janet is the third driver, if we assume that there are 3 drivers (other 2 being X and Y), the possible cases could be - J-X-Y and J-Y-X, where in the first case X comes 2nd and in the second case Y comes 2nd.

In this case probability would be

Event I: JXY - (1/5)(4/5)(4/5) = 16/125
Event II: JYX - (1/5)(4/5)(4/5) = 16/125

Probability = Event I + Event II = 16+16/125 = 32/125. Where am i going wrong?

Hi,
we are just interested in the third dive being perfect and for that the first two have to be imperfect dive..
it doesnt matter who dives first and who dives second..
you have taken X and Y, but there can be Z, T etc , after all its probability..

imperfect dive =4/5..
so third dive to be perfect.. 4/5*4/5*1/5=16/125
ans D
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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
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Bunuel wrote:
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.)

A. 1/5
B. 1/15
C. 4/25
D. 16/125
E. 61/125

Kudos for a correct solution.

Probability of perfect dive=1/5 and Imperfect dive=4/5
Probability that Janet gets a perfect score is when the 1st and 2nd diver have an imperfect score
Therefore, required probability=4/5*4/5*1/5=16/125
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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
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Bunuel wrote:
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.)

A. 1/5
B. 1/15
C. 4/25
D. 16/125
E. 61/125

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

In order for Janet to receive a perfect score, neither of the previous two divers can receive one. Therefore, you are finding the probability of a chain of three events: that diver one will not get a perfect score AND diver two will not get a perfect score AND Janet will get a perfect score. Multiply the probabilities: 4/5 x 4/5 x 1/5 = 16/125.

The probability is 16/125 that Janet will receive a perfect score.

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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
Bunuel wrote:
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.)

A. 1/5
B. 1/15
C. 4/25
D. 16/125
E. 61/125

Kudos for a correct solution.

I'm not sure why they worded the question this way. As I'd understand the question, as presented, the only answer I could give is "greater than 16/125". Janet certainly has a (4/5)(4/5)(1/5) = 16/125 probability of getting a perfect score on her first dive. But as they've worded the question, it sounds like each diver will dive multiple times - the comment in brackets suggests the divers will continue diving over the course of several "turns", and the first part of the question suggests that someone "will" get a perfect score. So I'd interpret that to mean the competition will continue until someone gets a perfect score, and that means Janet will potentially have many chances at a perfect dive. We'd need to know how many divers there are in total to then give an exact answer to the question. If there were three divers, and they kept diving until someone got a perfect score, the answer would be

(4/5)(4/5)(1/5) + (4/5)(4/5)(4/5)(4/5)(4/5)(1/5) + ...

= (4/5)^2 (1/5) + (4/5)^5 (1/5) + (4/5)^8 (1/5) + ...

= (16/125) (1 + (4/5)^3 + (4/5)^6 + ...)

The question is now beyond the scope of the GMAT -- that's an infinite geometric series in brackets, and you don't need to know any formulas for those. But using the geometric series formula, the sum in brackets adds to 1/(1 - 16/125) = 125/109, so the answer is 16/109 in this case. With more divers, the answer would be smaller, but still greater than 16/125.
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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
I have a question why isn't it this way

chance of 1st diver not making perfect dive is 4/5

chance of 2nd diver not making perfect dive is 3/4 as 1st diver already failed
and it is 1 among 4 to make perfect dive and 2nd diver not making perfect dive is 3/4
similarly 3rd diver making perfect dive is 1 among 3 which is 1/3

4/5*3/4*1/3

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Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
Bunuel wrote:
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.)

A. 1/5
B. 1/15
C. 4/25
D. 16/125
E. 61/125

Kudos for a correct solution.

3rd attempt is 1st Correct means other 2 are incorrect...

Incoorect Possibility*Possibility = Final Probability

ie, $$(1-\frac{1}{5})(1-\frac{1}{5})*\frac{1}{5} = \frac{16}{125}$$, Answer must be (D)
Re: In a diving competition, each diver has a 20% chance of a perfect dive [#permalink]
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