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Re: John will spend the summer in one of the houses either on [#permalink]
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There are 5 houses on Surf street ( 1L,4B) and 3 houses on Breakway Street (2L,1B). Now John can live in any one of the houses.
If he lives in Surf street , possible outcomes - 5 (1L,4B) .Therefore Probablity(Surfstreet) = 1/5
Similary, If he lives in Breakway street , possible outcomes - 3(2L,1B) , Probablity(Surfstreet) = 1/5

Hence 1/5 , 1/3 are possible probabilities depending upon where John lives.

Correct Option B and D
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Re: John will spend the summer in one of the houses either on [#permalink]
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sandy wrote:
Explanation

If you pare down the problem, then it is much simpler than it first appears. There will be 1 house on Surf Street at which a turtle will lay its eggs; this is “the number of outcomes you want.”

There are a total of 5 houses, which is the “total number of possible outcomes.” Therefore, if John spends the summer on Surf Street, the probability of John staying in the house where the sea turtle will lay its eggs is choice B, \(\frac{1}{5}\), and the information about the landside houses versus seaside houses actually turns out to be unnecessary.

Similarly, if John stays on Breaker Way, there will be 1 house that a turtle will lay its eggs at out of a total of 3 possible houses that John could live in. Therefore, if he lives on Breaker Way, there is a \(\frac{1}{3}\)chance that the turtle lays its eggs outsides John’s house, and this makes choice D correct.


The sea turtle will randomly lay eggs in front of any beach house in surf street. So, don't we multiply the probability of John lives in Surf Street (1/5) and the probability of sea turtle laying eggs in front beach house in Surf Street (1/4), because of the keyword randomly.
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Re: John will spend the summer in one of the houses either on [#permalink]
Carcass, sandy, the question states the following:
A sea turtle randomly lays eggs in front of a beachside house on Surf Street. There are 4 beachside house. Therefore, should we aim to determine in how many way a turtle should lay an egg in front of one of the 4 houses. 4C1 -total number of desirable outcomes; 5-total outcomes. --> Probability of laying eggs in front of 1 of the 4 houses=4C1/5. Where I am getting wrong with my approach?
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Re: John will spend the summer in one of the houses either on [#permalink]
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There are 5 houses on Surf street ( 1L,4B) and 3 houses on Breakway Street (2L,1B). Now John can live in any one of the house.
If he lives in Surf street , possible outcomes - 5 (1L,4B) .Therefore Probablity(Surfstreet) = 1/5
Similary, If he lives in Breakway street , possible outcomes - 3(2L,1B) , Probablity(Surfstreet) = 1/5

Hence 1/5 , 1/3 are possible probabilities depending upon where John lives.
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Re: John will spend the summer in one of the houses either on [#permalink]
Carcass, thank you for providing your response. According to the question, turtle lays eggs in front of Beachside houses on the SS. There are 4 and not 5 beachside houses on the SS. Therefore, should we have 4 and not 5 possible outcomes for houses on the SS? (The same question would apply for houses on the BS)
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Re: John will spend the summer in one of the houses either on [#permalink]
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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)..
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Re: John will spend the summer in one of the houses either on [#permalink]
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Re: John will spend the summer in one of the houses either on [#permalink]
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