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Bunuel
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I have one doubt here.

Question stem already says that the turtle lays its egg in front of the beach house. And Now question is asking the probability of laying the egg in front of the John's house. Therefore, we have to calculate the probability of living of John in Beach house where turtle will lay its egg.

From that point, only 4/5 and 1/3 is correct choice. 1/3 is in options. Therefore, only C is correct.
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I have one doubt here.

Question stem already says that the turtle lays its egg in front of the beach house. And Now question is asking the probability of laying the egg in front of the John's house. Therefore, we have to calculate the probability of living of John in Beach house where turtle will lay its egg.

From that point, only 4/5 and 1/3 is correct choice. 1/3 is in options. Therefore, only C is correct.

I think you missed the question " What are the probabilities that a sea turtle lays its eggs in front of John`s house, depending on which street he lives on? "

John can live in one of the beach house in one of the street. So our successful case will be 1 (1 beach house , either on Surf or Breakway street) always and possible cases will be sum of land and beach houses.

Hope it helps.
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John will spend the summer in one of the houses either on Surf Street or Breaker Way. Surf Street has 1 landside house and 4 beachside houses, and Breaker Way has 1 beachside house and 2 landside houses. A sea turtle randomly lays eggs in front of a beachside house on Surf Street, and another sea turtle randomly lays eggs in front of a beachside house on Breaker Way. What are the probabilities that a sea turtle lays its eggs in front of John`s house, depending on which street he lives on?

I. 1/5
II. 4/15
III. 1/3


A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

If John is staying at a house on Surf Street, then the probability is 4/5 x 1/4 = 1/5 that the turtle will lay eggs in front of John’s house (notice that 4/5 is the probability John will stay at a beachside house and 1/4 is the probability the turtle will lay eggs at the house in which John is staying).
If John is staying at a house on Breaker Way, then the probability is 1/3 x 1/1 = 1/1 that it will lay eggs in front of John’s house (notice that 1/3 is the probability John will stay at a beachside house and 1/1 is the probability the turtle will lay eggs at the house in which John is staying)..

Answer: D

Hi Scott,

I'm sorry, I still don't understand this, why are we not adding the two 'OR' probabilities here? I know that's not an answer option, but that's where I got stuck. Since John will EITHER live in Surf Street OR in Breaker Way.
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Bunuel
John will spend the summer in one of the houses either on Surf Street or Breaker Way. Surf Street has 1 landside house and 4 beachside houses, and Breaker Way has 1 beachside house and 2 landside houses. A sea turtle randomly lays eggs in front of a beachside house on Surf Street, and another sea turtle randomly lays eggs in front of a beachside house on Breaker Way. What are the probabilities that a sea turtle lays its eggs in front of John`s house, depending on which street he lives on?

I. 1/5
II. 4/15
III. 1/3


A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

If John is staying at a house on Surf Street, then the probability is 4/5 x 1/4 = 1/5 that the turtle will lay eggs in front of John’s house (notice that 4/5 is the probability John will stay at a beachside house and 1/4 is the probability the turtle will lay eggs at the house in which John is staying).
If John is staying at a house on Breaker Way, then the probability is 1/3 x 1/1 = 1/1 that it will lay eggs in front of John’s house (notice that 1/3 is the probability John will stay at a beachside house and 1/1 is the probability the turtle will lay eggs at the house in which John is staying)..

Answer: D

Hi Scott,

I'm sorry, I still don't understand this, why are we not adding the two 'OR' probabilities here? I know that's not an answer option, but that's where I got stuck. Since John will EITHER live in Surf Street OR in Breaker Way.


I believe we're not adding the two because the question says "depending on which street he stays"
i.e. The question is asking us to consider that he stays in one of the street, now calculate the probability that he is staying in a particular house of that street.
Am I right @buneul?
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Bunuel
John will spend the summer in one of the houses either on Surf Street or Breaker Way. Surf Street has 1 landside house and 4 beachside houses, and Breaker Way has 1 beachside house and 2 landside houses. A sea turtle randomly lays eggs in front of a beachside house on Surf Street, and another sea turtle randomly lays eggs in front of a beachside house on Breaker Way. What are the probabilities that a sea turtle lays its eggs in front of John`s house, depending on which street he lives on?

I. 1/5
II. 4/15
III. 1/3


A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

Probably a very silly thing point here. But we have 2 streets where john can live. So there will be 2 probabilities no matter what. Now only answer D has 2 probabilities.
All 3 cannot be and we cannot have one. If the options would have been A. I,II B. II,III C. I,III then it would have made sense to even calculate.?
Kindly help me understand.

P.S: Please assume I am thinking out loud.

Thank you.!
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TheNightKing
Both could have the same probability so A, B and C are in. I don't think 3 probabilities is an option though
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philipssonicare
TheNightKing
Both could have the same probability so A, B and C are in. I don't think 3 probabilities is an option though

Yeah I realised that later on. But still you can play your odds if you can't solve :) Just saying and not encouraging to go that route.!
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