Probability - Head & Tails..

Question

The probability is 1/2 that a certain coin will turn heads on any given toss. If a coin is to be tossed 4 times what is the probability of getting at least 2 tails?

Shouldn't the probability b same for heads and tails i.e (1/2) and hence the prob that all will be heads be 1/16..and then we follow the logic of 1-1/16

pls help me understand what am i doing wrong..

Raghava747 · Answer

The logic of doing it 1- the actual got when we have less than 2 tails would give us the answer. The ways in which we get less than two tails are {HHHH,HHHT,HHTH,HTHH,THHH} Hence 5 ways we get less than two tails. Now the total sample space is 16. To get two or more tails we simply can do 1-[5][/16] = 11/16

This would be same to get two or more heads..

Consider giving me kudos if the explanation was helpful. 8-)

DeeptiM · Answer

Thanks Raghava747..kudos to you!

pinchharmonic · Answer

I believe these are also valid ways to solve it. First thing I think of is that of the total set of 2^4 possibilities of 4 coin tosses, all of them are equally likely to occur. Meaning 4H is as likely as 4T, as well as 2H2T. This is because it's a 50% probability for HT.

To get 2 or more tails, then you could either have 2,3,4 tails.

4c2 + 4c3 + 4c4 
2 tails, 3 tails and 4 tails respectively
6 + 4 + 1 = 11

total possibilities = 2^4

11/16

or the possibilities of 0 tails and 1 tail would thus be all the combinations you don't want.

4c0 + 4c1 = 1 + 4 = 5
total possibilities = 2^4

1 - 5/16 = 11/16

Probability - Head & Tails..

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