The probability that the student will pass is; P(pass)
=
P(10) + P(9) + P(8) =
[10C10 + 10C9 + 10C8] / (.5)10 =
56/1024 ~
~ 0.0547 ~ 5.47%
where nCr =
n!/[r!(n-r)!]
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It is 0.3438
The difference between "at least" and "at most" is not restricted to probability. The difference is simply one between the precise meaning of the phrases in every day English language.
A fair chance of winning is more than 50% chance of winning. Therefore probability = 0.5 We need to find a fair chance of winning atleast one match. 1-(5/6)^n > 0.5 hence, n=4 QED (quite easily done, :-p)
Why do you think it is a question? It is a homework assignment.
powers of 10 are like this 10(small 1) 10(small 2) exetra i need help on the same question but atleast this might help someone