Probabilities of Unfair Coin Flipping

  • #1
I am having a difficult time on the following problem:

An unfair coin is flipped four times in a row. What is the probability of getting exactly two heads and two tails. The order does not matter as long as there are two head and two tails in the flip.

The probability of getting heads is P(H)=0.40
The probability of getting tails is P(T)=0.60

I tried this:

P(2H) = 4C2 * 0.40^4 = 0.1536

P(2T) = 4C2 * 0.60^4 = 0.7776

Which has me stumped because the probability of getting at least two heads should be 0.5248 according to my large tree diagram. What exactly is my P(2H) formula calculating? The P(2T) seems to correctly calculate the probability of getting at least two tails, according to my tree diagram. Any hints on how to solve this problem?
 

Answers and Replies

  • #2
D H
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What you are doing is wrong. Why are you taking [itex]4C2 P(H)^4[/itex] and [itex]4C2 P(T)^4[/itex]?

I don't know what your large tree diagram is. Your tree is right when it gives the probability of getting at least two heads is 0.5248. However, if your tree says the probability of getting at least two tails is 0.7776, it is wrong. The correct result (probability of at least two tails) is 0.8208.

What does the binomial distribution say about your problem?
 
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