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Mathematics 13 Online
OpenStudy (anonymous):

A fair coin is tossed 4 times in a row. If X is the number of heads counted, which table gives the probability distribution of X?

OpenStudy (anonymous):

@rational

OpenStudy (anonymous):

OpenStudy (rational):

whats the probability for getting heads on all 4 coins ?

OpenStudy (anonymous):

1/2?

OpenStudy (rational):

how ?

OpenStudy (rational):

may i know why do you think it is 1/2

OpenStudy (anonymous):

its a 50/50 chance to land on head.

OpenStudy (rational):

its a 50/50 chance to land head on one coin

OpenStudy (rational):

but when you have four coins, the probability for landing heads on "ALL" four coins need not be 1/2 right

OpenStudy (rational):

the probability is actually 1/2*1/2*1/2*1/2 = 1/16

OpenStudy (anonymous):

so what like 1/4?

OpenStudy (anonymous):

oh sorry. 4 coins. my bad

OpenStudy (rational):

Yes, another question whats the probability for getting NOT getting heads in any of the four coins ?

OpenStudy (anonymous):

1/2 each so total would be another 1/16?

OpenStudy (rational):

Yep! so P(X=0) is 1/16 and P(X=4) is also 1/16

OpenStudy (rational):

Now, whats the probability of getting exactly ONE head ?

OpenStudy (rational):

P(X=1) = ?

OpenStudy (anonymous):

I would think 1/2 but i know thats not it

OpenStudy (rational):

one way to get exactly ONE head is : H T T T

OpenStudy (rational):

notice that there are 3 other ways to get exactly ONE head : T H T T T T H T T T T H

OpenStudy (rational):

so the probability for getting exactly ONE head is 4/16 = 1/4

OpenStudy (rational):

see if that makes sense

OpenStudy (anonymous):

Yes it does, I see what you mean now. okay.

OpenStudy (rational):

look at the options

OpenStudy (anonymous):

A?

OpenStudy (rational):

|dw:1432817476450:dw|

OpenStudy (rational):

Yes! our work so far matches perfectly with option A

OpenStudy (anonymous):

Yes thats why I said A, the work matches A and not others

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