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#1 2024-05-13 10:23:21

mathxyz
Member
From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Trains to Coney Island

Four friends (Amy, Bobby, Chris and Dusty) each choose a random train to Coney Island between 1 and 5. What is the probability that any of them chose the same train?

I am lost here. A friend hinted that this is conditional probability. In what way is this conditional probability? How do I solve this problem?

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#2 2024-05-13 19:42:24

Keep_Relentless
Member
From: Queensland, Australia
Registered: 2024-05-05
Posts: 61

Re: Trains to Coney Island

Since there are a number of ways any of them could take the same train, I would instead do the probability that none of them chose the same train.

Amy picks a train: 5/5
Bobby picks a different train: 4/5
Chris picks yet another train: 3/5
Dusty picks another different train: 2/5

Multiply those together, and you get the probability they all chose a different train.

Take this probability away from 1, and you get the probability any of them chose the same train.

I think that's right?

K_R


"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

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#3 2024-05-14 00:06:40

Europe2048
Member
Registered: 2024-01-03
Posts: 38

Re: Trains to Coney Island

Remember: you can use the hide tag to make your answer hidden in a button.

Last edited by Europe2048 (2024-05-14 08:53:27)

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#4 2024-05-14 01:21:15

mathxyz
Member
From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Re: Trains to Coney Island

Europe2048 wrote:

Remember: you can use the hide tag to make your answer hidden in a button.

Thank you for your great effort but the textbook answer is 101/125.
Is the textbook wrong?

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#5 2024-05-14 01:22:32

mathxyz
Member
From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Re: Trains to Coney Island

Keep_Relentless wrote:

Since there are a number of ways any of them could take the same train, I would instead do the probability that none of them chose the same train.

Amy picks a train: 5/5
Bobby picks a different train: 4/5
Chris picks yet another train: 3/5
Dusty picks another different train: 2/5

Multiply those together, and you get the probability they all chose a different train.

Take this probability away from 1, and you get the probability any of them chose the same train.

I think that's right?

K_R

Thanks bit the textbook answer is 101/125.

Is the textbook wrong? I must admit that this one is a tricky question.

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#6 2024-05-14 01:34:31

Keep_Relentless
Member
From: Queensland, Australia
Registered: 2024-05-05
Posts: 61

Re: Trains to Coney Island

It is 101/125, Europe just accidentally wrote 101/25.

K_R


"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

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#7 2024-05-14 06:16:36

mathxyz
Member
From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Re: Trains to Coney Island

Keep_Relentless wrote:

It is 101/125, Europe just accidentally wrote 101/25.

K_R

Thank you for the information.

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#8 2024-05-14 08:54:08

Europe2048
Member
Registered: 2024-01-03
Posts: 38

Re: Trains to Coney Island

Sorry for the mistake, I fixed it.

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#9 2024-05-14 11:01:24

mathxyz
Member
From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Re: Trains to Coney Island

Europe2048 wrote:

Sorry for the mistake, I fixed it.

Very cool. Thanks.

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