goneewindy wrote:

Does anyone has a faster way of solving this problem instead of drawing out the tree?

A fair coin is flipped 5 times.

Quantity A: The probability of getting more heads than tails

Quantity B: 1/2

There are 32 sample solutions in the solution set of the 5 coin toss. Since \(2^5 = 35\)

Now there are 5 coins so number of Heads can either be greater than or less than Tails. Since each head or tail is equally likely the probability of getting more heads is 0.5 or 16/32. The sample space is divided in two groups, "heads are more" and "tails are more".

If there had been even number of coins say 6. The sample space would have been 64. Now it wold have been divided into 3 groups, "heads are more" "tails are more" and "heads are equal to tails".

Now the "equal" case has 20 solutions. \(6!/3!*3!\) =20. The remaining 44 sample space points are equally likely to have heads or tails more.

Don't use logic unless you are well versed with probability theory. It is always easier and safer to draw the tree if you have the time.
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Sandy

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