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
So that's the first problem, and here's the OE:
But here's another problem that's identical and is solved completely differently:
Molly is rollling a number cube with faces numbered 1 to 6 repeatedly. When she receives a 1, she will stop rolling the cube. What is the probability that Molly will roll the die less than 4 times before stopping?
A) 1/6
B) 5/36
C)1/216
D)91/216
E) 1/2
So here's what I don't get:
In the first problem you only calculated the odds of the diver NOT getting a perfect score, and in the second one you calculated the odds of NOT getting a 1 and then added to it the odds of not getting a 1, multiplied by the odds of getting something else, and then added to that the odds of not getting a 1 again, and multiplied it by the odds of rolling something else twice.
Using that logic the first problem should have been solved by multiplying the odds of the first diver getting a perfect score (1/5) and then adding to that the odds of the 2nd diver getting a perfect score, multiplied by their odds of getting something else (1/5)(4/5) and then adding to that the odds that the first and second divers don't get a perfect score, multiplied by the odds that she does get a perfect score (1/5)(4/5)(4/5), so the answer should be 61/125.![]()
A) 1/5
B) 1/15
C) 4/25
D) 16/125
E) 61/125
So that's the first problem, and here's the OE:
[Reveal] Spoiler: OE
D. 16/125: 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.
But here's another problem that's identical and is solved completely differently:
Molly is rollling a number cube with faces numbered 1 to 6 repeatedly. When she receives a 1, she will stop rolling the cube. What is the probability that Molly will roll the die less than 4 times before stopping?
A) 1/6
B) 5/36
C)1/216
D)91/216
E) 1/2
[Reveal] Spoiler: OE
The setup for this problem is actually more complex than it may initially appear. First, you need to separate three possibilities from each other. If Molly rolls less than 4 times before stopping, then she rolls a 1 on her first roll OR on her second roll OR on her third roll. That means that you will ultimately have to ADD the three probabilities:
P(lst roll end) + P(2nd roll) + P(3rd roll)
The probability of rolling a 1 on her first roll is 1/6. P(lst roll end) = 1/6. If Molly does not roll a 1 until her second roll, that means she did not roll a 1 on her first roll. So to get
P(2nd roll end), you need to calculate the probability of not rolling a 1 on the first roll AND rolling a 1 on the second roll.
P(2nd roll end) = P(Not 1 on 1st) x P(1 on 2nd)
Similarly, if Molly does not roll a 1 until her third roll, you need the probability of not 1 on her first roll AND not 1 on her second roll AND 1 on her third roll:
P(3rd roll end) = P(Not 1 on 1st) x P(Not 1 on 2nd) x P(1 on 3rd)
The probability that she rolls a 1 on her first roll is 1/6. There are 6 possible outcomes, but only 1 of them is desirable. The probability of not rolling a 1 is then equal to 1 minus the probability of rolling a 1.1- 1/6 = 5/6. You can fill in the appropriate probabilities and solve:
P(lst roll) + P(2nd roll) + P(3rd roll) = (1/6) + (1/6)(5/6)+(1/6)(5/6)(5/6)= 91/216
P(lst roll end) + P(2nd roll) + P(3rd roll)
The probability of rolling a 1 on her first roll is 1/6. P(lst roll end) = 1/6. If Molly does not roll a 1 until her second roll, that means she did not roll a 1 on her first roll. So to get
P(2nd roll end), you need to calculate the probability of not rolling a 1 on the first roll AND rolling a 1 on the second roll.
P(2nd roll end) = P(Not 1 on 1st) x P(1 on 2nd)
Similarly, if Molly does not roll a 1 until her third roll, you need the probability of not 1 on her first roll AND not 1 on her second roll AND 1 on her third roll:
P(3rd roll end) = P(Not 1 on 1st) x P(Not 1 on 2nd) x P(1 on 3rd)
The probability that she rolls a 1 on her first roll is 1/6. There are 6 possible outcomes, but only 1 of them is desirable. The probability of not rolling a 1 is then equal to 1 minus the probability of rolling a 1.1- 1/6 = 5/6. You can fill in the appropriate probabilities and solve:
P(lst roll) + P(2nd roll) + P(3rd roll) = (1/6) + (1/6)(5/6)+(1/6)(5/6)(5/6)= 91/216
So here's what I don't get:
In the first problem you only calculated the odds of the diver NOT getting a perfect score, and in the second one you calculated the odds of NOT getting a 1 and then added to it the odds of not getting a 1, multiplied by the odds of getting something else, and then added to that the odds of not getting a 1 again, and multiplied it by the odds of rolling something else twice.
Using that logic the first problem should have been solved by multiplying the odds of the first diver getting a perfect score (1/5) and then adding to that the odds of the 2nd diver getting a perfect score, multiplied by their odds of getting something else (1/5)(4/5) and then adding to that the odds that the first and second divers don't get a perfect score, multiplied by the odds that she does get a perfect score (1/5)(4/5)(4/5), so the answer should be 61/125.