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A fair coin is tossed 8 times. What is the probability of obtaining exactly 4 heads if it is known that at least 1 head appeared? Round your answer to 4 decimal places
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If, in the sequence of 8 tossings, that 1 head already appeared and counted as one of the 4? (Just clarifying.)
This is sort of a trick question. Stating that 1 head appeared just means that in the subsequent 7 tosses, what are the odds that you will get 3 more heads?
If thats the case then i would want 3 successes in a number of 7 trials now with the probability of 1/2
Bayes rulez! (C(7,3)*0.5^7)/(1-0.5^8)...round that urself
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