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Probability-Tossing coins

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Probability-Tossing coins

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Is there any other method to solve this apart from listing down the options?

Urpi and Manco are playing a game where they flip a fair coin four times and try to predict the outcomes.

What is P(A), the probability that the second flip is heads?

What is P(B), the probability that the fourth flip is heads?

What is P(A and B), the probability that the second flip is heads and the fourth flip is heads?

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Certainly!

For P(A), you have 2 * 2 * 2 * 2 = 16 total options. Your target is "Heads 2nd", for which there are 2 * 1 * 2 * 2 = 8 total options, giving you 8 / 16.

For P(B), same thing.

For P(A and B), you have 2 * 1 * 2 * 1 = 4 target options, for 1/4.

P(A and B) here = P(A) * P(B), so you could just do 1/2 * 1/2 as well.

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Hi ash4gmat,

As complex as this situation might seem, the 'math' behind it is rather simple. Coin flips are independent events - the result of one flip has NO impact on the results of any other flips. Thus, the probability of ANY flip being 'heads' is the same: 1/2.

These types of questions become more layered if you're asked to calculate a more complex outcome. The probability of flipping TWO heads on two tosses can be solve with a little table or a bit of math.

HH
HT
TH
TT

Here, there are only 4 possible outcomes with one of them matching what we're looking for. So the probability is 1/4. Mathematically, we can also solve it this way:

Probability of heads on the first flip: 1/2
Probability of heads on the second flip: 1/2
Probability of BOTH events occurring: (1/2)(1/2) = 1/4

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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