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Sagot :
To asnwer this questions we can use the binomial distribution. The probability of having a number k of successes in a binomial experiment is given by:
[tex]P(X=k)=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}[/tex]where n is the number of trials and p is the probability of succes.
a.
Since we like to have a one in each roll this means that in this case the probability of succes will be 1/6 (1 possibility out of 6). Also we have 8 rolls then n=8, and we like that the one is the result in each of them, then k=8. Plugging this values in the distribution we have:
[tex]\begin{gathered} P(X=8)=\frac{8!}{8!(8-8)!}(\frac{1}{6})^8(1-\frac{1}{6})^{8-8} \\ =\frac{1}{1679616} \\ =0.595\times10^{-6} \end{gathered}[/tex]Therefore the probability of getting a one in each roll is 0.00000595.
b.
Since we like a 6 exactly twice this means that k=2. The probability of succes is 1/6. Plugging the values in the distribution we have:
[tex]\begin{gathered} P(X=2)=\frac{8!}{2!(8-2)!}(\frac{1}{6})^2(1-\frac{1}{6})^{8-2} \\ =0.26 \end{gathered}[/tex]Therefore the probability of obtaining 6 exactly twice is 0.26.
c.
The probability of obtaining at least once a six is the sum of obtaining 1 and obtaining 2 and obtaining 3 and so on.
That means that the probability is:
[tex]\begin{gathered} P=P(X=1)+P(X=2)+P(X=3)+P(X=4) \\ +P(X=5)+P(X=6)+P(X=7)+P(X=8) \end{gathered}[/tex]but this is more easily obtain if we notice that this is the same as:
[tex]P=1-P(X=0)[/tex]This comes from the fact that the sum of all the successes possibilities (in this case obtaining a 6) have to be 1.
Then the probability of obtaniing at least once a six is:
[tex]\begin{gathered} P=1-P(X=0) \\ =1-\frac{8!}{0!(8-0)!}(\frac{1}{6})^0(1-\frac{1}{6})^{8-0} \\ =0.767 \end{gathered}[/tex]Therefore the probability of obtaining at least once a six is 0.767.
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