Another Probability Question!

[n2o*]

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With one fair die, find the probability of throwing two fours in five attempts.

The answer is (10*125)/ (36*216)

Explain pls.

 

[786mine]

I think the answer should be 10.

here is my logic:

n=5 (total # of throws)
k=2 (total # of number we want)
formula

n!/(k!(n-k)!) = 5! / ( 2! (5-2)!) = 10

 

[organichemistry]

I think the answer should be 10.

here is my logic:

n=5 (total # of throws)
k=2 (total # of number we want)
formula

n!/(k!(n-k)!) = 5! / ( 2! (5-2)!) = 10

the probability can't be 10.

probability has to be on the interval [0, 1].

10 tells you how many different ways you could throw the die 5 times and get 4 twice.


here's how i would do it:

P(X=2) = (5 choose 2) * (1/6)^2 * (5/6)^3, where X=2 "successes"
P(X=2) = 10 * 1/36 * 125/216
P(X=2) = 0.16075

...which I believe is equivalent to that solution that the OP posted.

remember... 5 choose 2... because you are throwing it 5 times and are "choosing" 4 twice. 1/6 to the second because there is a 1 in 6 chance of rolling a 4, and you have to do it exactly twice. 5/6 to the third, because there is a 5 in 6 chance of NOT rolling a 4, and you have to do it exactly 3 times.

hope this helps!

 

[786mine]

the probability can't be 10.

probability has to be on the interval [0, 1].

10 tells you how many different ways you could throw the die 5 times and get 4 twice.


here's how i would do it:

P(X=2) = (5 choose 2) * (1/6)^2 * (5/6)^3, where X=2 "successes"
P(X=2) = 10 * 1/36 * 125/216
P(X=2) = 0.16075

...which I believe is equivalent to that solution that the OP posted.

remember... 5 choose 2... because you are throwing it 5 times and are "choosing" 4 twice. 1/6 to the second because there is a 1 in 6 chance of rolling a 4, and you have to do it exactly twice. 5/6 to the third, because there is a 5 in 6 chance of NOT rolling a 4, and you have to do it exactly 3 times.

hope this helps!

you are right. i didn't read the question properly. i was thinking of something else. sorry.

 

[n2o*]

the probability can't be 10.

probability has to be on the interval [0, 1].

10 tells you how many different ways you could throw the die 5 times and get 4 twice.


here's how i would do it:

P(X=2) = (5 choose 2) * (1/6)^2 * (5/6)^3, where X=2 "successes"
P(X=2) = 10 * 1/36 * 125/216
P(X=2) = 0.16075

...which I believe is equivalent to that solution that the OP posted.

remember... 5 choose 2... because you are throwing it 5 times and are "choosing" 4 twice. 1/6 to the second because there is a 1 in 6 chance of rolling a 4, and you have to do it exactly twice. 5/6 to the third, because there is a 5 in 6 chance of NOT rolling a 4, and you have to do it exactly 3 times.

hope this helps!

it helped a lot! thank u so much 🙂

 

[seoul]

10 tells you how many different ways you could throw the die 5 times and get 4 twice.

remember... 5 choose 2... because you are throwing it 5 times and are "choosing" 4 twice.

The permutation formula posted by 786mine: n!/(k!(n-k)!) = 5! / ( 2! (5-2)!) = 10 will better explain for the 10 ways of throwing the dice here.