If a fair coin is tossed four times, what is the probability of obtaining at most one head?
I know the answer is 5/16 but how do you get that? Please Help!
Stacey Warren - Expert brainly.com
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The formula for n "successes" out of m tries is
n Choose m p^n * q^(n-m)
where p is the probability of success and q is 1-p (probability of failure)
in your case
4C1 * (1/2)^1 * (1/2)^3
4 * 1/2* 1/8 = 4/16
at most one head means 1 head or 0 heads
the chance for 0 heads is
4C0 * (1/2)^0 * (1/2)^4
1 * 1 * 1/16
so the chance of 1 head or 0 heads is 4/16 + 1/16 = 5/16
thank you- why are you multiplying 4C1 by (1/2)^1 and (1/2)^3
you could work it out in detail, by saying
the chance of 0 heads is TTTT
1/2 chance of 0 heads on the first toss
times 1/2 chance of heads on the 2nd toss, etc to get
1/2 * 1/2 * 1/2 * 1/2 = 1/16 chance of no heads
the chance of 1 head has 4 cases:
1/2 chance of heads on the first toss times
1/2 chance of tails on the 2nd toss * 1/2 chance of tails on 3rd toss * 1/2 chance of tails on the 4th toss = 1/2 *1/2 * 1/2 * 1/2 = 1/6
you can work out that each has a chance of 1/16
all together you get 4/16 chance of 1 head and 1/16 chance of 0 heads,
or 5/16 chance of 0 or 1 head
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Oh I get it! thank you so much!
the 4C1 is the count of the ways of getting one head in 4 tries. The 4 combinations are showed up above.
It gets more complicated for say, 3 heads in 5 tries.
you can work out the combinations, but there will be 5C3 of them