1992 AHSME Problems/Problem 29
Contents
Problem
An unfair coin has a probability of turning up heads. If this coin is tossed times, what is the probability that the total number of heads is even?
Solution
Doing casework on the number of heads (0 heads, 2 heads, 4 heads...), we get the equation This is essentially the expansion of but without the odd power terms. To get rid of the odd power terms in , we add and then divide by because the even power terms that were not canceled were expressed twice. Thus, we have Or which is equivalent to answer choice .
Solution 2
Denote as the probability that the number of heads is even. Then we have two cases:
: (There are an odd number of heads for coin flips). Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for flips?" That is precisely . Now, since we want an even number of heads, what is the probability that on the coin flip we get a head? . So we have .
: (There are an even number of heads for coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for flips will be odd. :/ So what is the probability we get a tail? . Thus we have .
Adding both cases, we have that . Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in .
Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D
(Or just come up with the general formula Simple observation really. If you really want to get mathy, prove this through induction).
~th1nq3r
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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