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Probability = Sea Turtle lays eggs in front of Jhon's House?
[#permalink]
Updated on: 12 Jun 2019, 10:09

1

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Question Stats:

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?

Select all that apply

A). \(\frac{4}{25}\)

B). \(\frac{1}{5}\)

C). \(\frac{4}{15}\)

D). \(\frac{1}{3}\)

E). \(\frac{17}{15}\)

Select all that apply

A). \(\frac{4}{25}\)

B). \(\frac{1}{5}\)

C). \(\frac{4}{15}\)

D). \(\frac{1}{3}\)

E). \(\frac{17}{15}\)

Re: Probability = Sea Turtle lays eggs in front of Jhon's House?
[#permalink]
13 Aug 2021, 16:16

2

The problem is asking what the different probabilities would be for each street he could choose to move into.

This problem naturally breaks down into two cases:

Case 1: John moves into a house in Surf Street

1. The sea turtle only lays eggs in front of beachside houses. Hence, John would have to move into a beachside house.

2. That means he has a 4/5 chance of moving into a beachside house on that street.

3. Then, once he's moved in, the turtle would have to choose his beachside house to lay an egg in front of; it has four choices, so the odds of that happening are 1 in 4.

4. For two independent events, the probability of them both occurring is just the probability of the first one times the probability of the second once. Hence, (4/5)(1/4) = 4/20 = 1/5.

Case 2: John moves into a house on Breaker Way

1. John still needs to move into a beachside house for there to even be a possibility of him encountering an egg, so this is 1/3 in this case.

2. The turtle only has one beachside house to choose from to lay eggs in front of, so it's a 1/1 chance it will lay an egg in front of that house.

3. Again, we multiply together to get 1/3.

From the options given above, B and D equal 1/5 and 1/3 respectively.

This problem naturally breaks down into two cases:

Case 1: John moves into a house in Surf Street

1. The sea turtle only lays eggs in front of beachside houses. Hence, John would have to move into a beachside house.

2. That means he has a 4/5 chance of moving into a beachside house on that street.

3. Then, once he's moved in, the turtle would have to choose his beachside house to lay an egg in front of; it has four choices, so the odds of that happening are 1 in 4.

4. For two independent events, the probability of them both occurring is just the probability of the first one times the probability of the second once. Hence, (4/5)(1/4) = 4/20 = 1/5.

Case 2: John moves into a house on Breaker Way

1. John still needs to move into a beachside house for there to even be a possibility of him encountering an egg, so this is 1/3 in this case.

2. The turtle only has one beachside house to choose from to lay eggs in front of, so it's a 1/1 chance it will lay an egg in front of that house.

3. Again, we multiply together to get 1/3.

From the options given above, B and D equal 1/5 and 1/3 respectively.

Re: Probability = Sea Turtle lays eggs in front of Jhon's House?
[#permalink]
Updated on: 24 Aug 2016, 04:36

We can construct a table from the information given in this question:

--------------------------------------------------------------------------------------------------------

Surf Street | Beaker Way

--------------------------------------------------------------------------------------------------------

1 Landside house (Turtle lays egg before it) | 2 Land-side houses

--------------------------------------------------------------------------------------------------------

4 Beach-house houses | 1 Beach-house house (Turtle lays egg before it)

--------------------------------------------------------------------------------------------------------

What does randomly here mean?

How can we construct a table in user interface of this community for posting a reply or creating a new post?

--------------------------------------------------------------------------------------------------------

Surf Street | Beaker Way

--------------------------------------------------------------------------------------------------------

1 Landside house (Turtle lays egg before it) | 2 Land-side houses

--------------------------------------------------------------------------------------------------------

4 Beach-house houses | 1 Beach-house house (Turtle lays egg before it)

--------------------------------------------------------------------------------------------------------

What does randomly here mean?

How can we construct a table in user interface of this community for posting a reply or creating a new post?

Re: Probability = Sea Turtle lays eggs in front of Jhon's House?
[#permalink]
24 Aug 2016, 04:35

2

Probability of John to live in a house before which a turtle lays eggs : Probability of John's living in Land-side house on surf Street +Probability of John living in Beach-side house on Beaker Way = P(John in Land-side house on surf Street) + P(John in Beach-side house on Beaker Way)

= (1/5)+(1/3) = 8/15

We are actually required to calculate the possible probabilities that sea turtle lays eggs in front of John's house depending upon which street he lives on.

Now suppose:

\(Pjls\) = Probability of John living in Land-side house on surf Street

\(Pjbs\) = Probability of John living in Beach-side house on Beaker Way

\(Ptls\) = Probability of turtle laying egg before Land-side house on surf Street

\(Ptbs\) = Probability of turtle laying egg before Beach-side house on Beaker Way

\(Ptej\) = Probability of turtle laying egg before house in which John lives

\(Pjls\) = 1/5

\(Pjbs\) = 1/3

\(Ptej\) = (1/5) * \(Ptls\) + (1/3) * \(Ptbs\)

What I am able to understand is that if the turtle lays egg randomly it means the probability of each turtle laying egg before a specific house varies between 0 and 1 i.e. it either lays eggs or it does not lay egg;in other words its probability only has two values 1 or 0.

If first turtle lays egg and second does not lay egg then

\(Ptej\) = (1/5) * 1 + (1/3) * 0 = 1/5 ... a

If 2nd turtle lays egg and first does not lay egg then

\(Ptej\) = (1/5) * 0 + (1/3) * 1 = 1/3 ... b

Other probabilities could be

\(Ptej\) = (1/5) * 1 + (1/3) * 1 = 8/15

\(Ptej\) = (1/5) * 0 + (1/3) * 0 = 0

Since Equations (a) and (b) give answer choices that give us answer choices from given list so these are the required answers

Could anybody endorse the solution to this question that I have calculated?

I have edited this post to provide a complete solution

= (1/5)+(1/3) = 8/15

We are actually required to calculate the possible probabilities that sea turtle lays eggs in front of John's house depending upon which street he lives on.

Now suppose:

\(Pjls\) = Probability of John living in Land-side house on surf Street

\(Pjbs\) = Probability of John living in Beach-side house on Beaker Way

\(Ptls\) = Probability of turtle laying egg before Land-side house on surf Street

\(Ptbs\) = Probability of turtle laying egg before Beach-side house on Beaker Way

\(Ptej\) = Probability of turtle laying egg before house in which John lives

\(Pjls\) = 1/5

\(Pjbs\) = 1/3

\(Ptej\) = (1/5) * \(Ptls\) + (1/3) * \(Ptbs\)

What I am able to understand is that if the turtle lays egg randomly it means the probability of each turtle laying egg before a specific house varies between 0 and 1 i.e. it either lays eggs or it does not lay egg;in other words its probability only has two values 1 or 0.

If first turtle lays egg and second does not lay egg then

\(Ptej\) = (1/5) * 1 + (1/3) * 0 = 1/5 ... a

If 2nd turtle lays egg and first does not lay egg then

\(Ptej\) = (1/5) * 0 + (1/3) * 1 = 1/3 ... b

Other probabilities could be

\(Ptej\) = (1/5) * 1 + (1/3) * 1 = 8/15

\(Ptej\) = (1/5) * 0 + (1/3) * 0 = 0

Since Equations (a) and (b) give answer choices that give us answer choices from given list so these are the required answers

Could anybody endorse the solution to this question that I have calculated?

I have edited this post to provide a complete solution

Re: Probability = Sea Turtle lays eggs in front of Jhon's House?
[#permalink]
28 Aug 2016, 20:55

Dear Mr. Brent and Sandy please endorse or rectify the solution to this question so that concepts may be cleared

Re: Probability = Sea Turtle lays eggs in front of Jhon's House?
[#permalink]
07 Apr 2018, 09:26

1

The probability of John spending the summer in Surf Street = 1/2

The probability of John spending the summer in Beaker Way = 1/2

The probability a sea turtle lays eggs in front of John's house in Surf Street = 1/5 * 1/2 = 1/10

The probability a sea turtle lays eggs in front of John's house in Beaker Way= 1/3 * 1/ 2 = 1/6

The probability a sea turtle lays eggs in front of John's house is the sum of the above = 16/60 or 4/15

The answer is C

The probability of John spending the summer in Beaker Way = 1/2

The probability a sea turtle lays eggs in front of John's house in Surf Street = 1/5 * 1/2 = 1/10

The probability a sea turtle lays eggs in front of John's house in Beaker Way= 1/3 * 1/ 2 = 1/6

The probability a sea turtle lays eggs in front of John's house is the sum of the above = 16/60 or 4/15

The answer is C

Re: Probability = Sea Turtle lays eggs in front of Jhon's House?
[#permalink]
12 Jun 2019, 05:46

need elaboration pls

Re: Probability = Sea Turtle lays eggs in front of Jhon's House?
[#permalink]
12 Jun 2019, 10:12

2

Expert Reply

1

Bookmarks

Living in S Street is where people choose one in five houses. Turtles choose one in four houses. There are four cases in which they correspond.

So \(\frac{4}{(5*4)}=\frac{1}{5}\)

Stay at B Street, choose one of the three houses, and the turtle has only one choice.

So \(\frac{1}{(1*3)}=\frac{1}{3}\)

_________________

So \(\frac{4}{(5*4)}=\frac{1}{5}\)

Stay at B Street, choose one of the three houses, and the turtle has only one choice.

So \(\frac{1}{(1*3)}=\frac{1}{3}\)

_________________

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Probability = Sea Turtle lays eggs in front of Jhon's House?
[#permalink]
07 Jul 2021, 20:01

2

John has the option to choose the houses on the Surf Street and the Breaker Way. We are asked of the individual probabilities depending on the street that John has chosen to live.

If John has chosen the Surf Street then he has 5 choices. But the sea turtle only lay eggs in front of the beachside house. There are 4 of them here.

So John can choose 4 houses out of four if he wants that sea turtle lay eggs in front of it.

So 4/5.

For the turtle there are 4 choices of the houses(beachside). But they can only choose 1 from it, the one that John has chosen already.

That is probabilities that a sea turtle lays its eggs in front of John house is (4/5)*(1/4) that is 1/5.

For the Breaker Way, for John there are 3 houses out of which only one is beachside. So 1/3.

For the turtle there is only one choice of the houses and they have to choose that one only. So 1/1.

That is probabilities that a sea turtle lays its eggs in front of John house is (1/3)*(1/1) that is 1/3.

So B,D.

If John has chosen the Surf Street then he has 5 choices. But the sea turtle only lay eggs in front of the beachside house. There are 4 of them here.

So John can choose 4 houses out of four if he wants that sea turtle lay eggs in front of it.

So 4/5.

For the turtle there are 4 choices of the houses(beachside). But they can only choose 1 from it, the one that John has chosen already.

That is probabilities that a sea turtle lays its eggs in front of John house is (4/5)*(1/4) that is 1/5.

For the Breaker Way, for John there are 3 houses out of which only one is beachside. So 1/3.

For the turtle there is only one choice of the houses and they have to choose that one only. So 1/1.

That is probabilities that a sea turtle lays its eggs in front of John house is (1/3)*(1/1) that is 1/3.

So B,D.

gmatclubot

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