If 4 coins are tossed, find the probabilities

[rowdy3]

If 4 coins are tossed, find the following probability:

2 heads.

more than 3 tails.

For 2 heads I got 1/16.
More than 3 heads I don't know how to start that problem.

 

[Galadirith]

Hi rowdy3,

First I need to ask do you mean "more that 3 heads" or "more than 3 tails" as you wrote both in you question :D, ill assume for now that you actually meant more than 3 heads.

So you two probabilities would be:

P₁ : exactly 2 heads
P₂ : more than 3 heads (this could also be written "exactly 4 heads"!)

now first how did you work out exactly 2 heads was 1/16. Show us how you got to that answer , is that in fact correct? then we can help you further :D

Are you using a probability tree, as if you are trying to work out the probability by thinking about it, you may get the incorrect answer, a probability tree will help you visualise the question much easier.

 

[rowdy3]

I did it wrong. I should do 4c2/ 2^4 which will come out to 6/16 for 2 heads.
More than 3 tails I don't know how.

 

[Mark44]

With 4 coins, "more than 3 tails" means "3 tails" or "4 tails." Can you find the probability of each? Since these are independent events, their probabilities add.

 

[rowdy3]

more than 3 heads.

odds of 1 coin landing on tails 1/2
odds of 2 coins landing on tails 1/2 * 1/2 = 1/4
odds of 3 coins tails 1/2 * 1/2 * 1/2 = 1/8
odds of 4 coins tails 1/2 * 1/2 * 1/2* 1/2 = 1/16

 

[lanedance]

more than 3 heads.

odds of 1 coin landing on tails 1/2
odds of 2 coins landing on tails 1/2 * 1/2 = 1/4
odds of 3 coins tails 1/2 * 1/2 * 1/2 = 1/8
odds of 4 coins tails 1/2 * 1/2 * 1/2* 1/2 = 1/16

not quite - i would re-write what you have as
T - odds of 1^st coin landing on tails 1/2
TT - odds of 1^st & 2^nd coins landing on tails 1/2 * 1/2 = 1/4
TTT - odds of 1^st & 2^nd & 3^rd coins tails 1/2 * 1/2 * 1/2 = 1/8
and so on...

now consider this case when the coin is tossed 4 times, with only the last toss being a tail, ie.
HHHT
the probability of this event (1/2)^4 = 1/16

Any individual ordered outcome will have a discrete probability of 1/16, as there are 16 different ordered outcomes...
so there are 4 different ways to get a single tail
THHH, HTHH, HHTH, HHHT
so the total probability of a single head will be 4*(1/16) = 1/4

so for the case where you want to find 2 heads, you must sum the probability of all the cases where you have two heads... eg.
HHTT, HTHT, HTTH...
any good ideas on how to count the cases?

 

[rowdy3]

I'm sorry, it's more than 3 tails. Would it still be 1/16?

 

[lanedance]

no, any single outcome has probability of 1/16,

you need to count all the possible outcomes with 3 or 4 tails...

(note that it may be easier to consider the cases of 1 or less heads, which is equivalent)

 

[Galadirith]

Hi guys, I thought Id comment again, surly this is wrong, if the probability was:

"Find the probability of 3 or more tails" (thanks rowdy for clarify which ;D)

then yes it would be the probability of exactly 3 heads (all permutations) and exactly 4 heads (only one permutation), but its not the probability as rowdy wrote it is:

"Find the probability of more than 3 tails"

which is exactly 4 tails, NOT 3 tails (all permutations), in which case rowdy you would be correct with the probability being 1/16, am I crazy or something :D But rowdy awesome on the first one, well done for spotting the mistake.

 

[frustr8photon]

a) 6/16
b) 1/16