Question

1. you flip a fair coin 8 times. let ( x = ) the number of tails that come up. find:

a. ( p(x = 4) )
b. ( p(x lt 2) )
c. ( p(x gt 4) )
d. ( p(x = 8) )
e. ( p(x leq 5) )
f. ( p(x geq 7) )

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here \(n = 8\), \(p=\frac{1}{2}\) (since it's a fair coin).

Step2: Calculate \(P(X = 4)\)

For \(A\), \(n = 8\), \(k = 4\), \(p=\frac{1}{2}\).
\(C(8,4)=\frac{8!}{4!(8 - 4)!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70\)
\(P(X = 4)=C(8,4)\times(\frac{1}{2})^{4}\times(\frac{1}{2})^{8 - 4}=70\times(\frac{1}{2})^{8}=\frac{70}{256}=\frac{35}{128}\approx0.273\)

Step3: Calculate \(P(X\lt2)\)

For \(B\), \(P(X\lt2)=P(X = 0)+P(X = 1)\)
\(P(X = 0)=C(8,0)\times(\frac{1}{2})^{0}\times(\frac{1}{2})^{8}=\frac{1}{256}\)
\(C(8,1)=\frac{8!}{1!(8 - 1)!}=8\)
\(P(X = 1)=C(8,1)\times(\frac{1}{2})^{1}\times(\frac{1}{2})^{7}=\frac{8}{256}\)
\(P(X\lt2)=\frac{1 + 8}{256}=\frac{9}{256}\approx0.035\)

Step4: Calculate \(P(X\gt4)\)

For \(C\), \(P(X\gt4)=P(X = 5)+P(X = 6)+P(X = 7)+P(X = 8)\)
\(C(8,5)=\frac{8!}{5!(8 - 5)!}=56\), \(P(X = 5)=56\times(\frac{1}{2})^{8}\)
\(C(8,6)=\frac{8!}{6!(8 - 6)!}=28\), \(P(X = 6)=28\times(\frac{1}{2})^{8}\)
\(C(8,7)=\frac{8!}{7!(8 - 7)!}=8\), \(P(X = 7)=8\times(\frac{1}{2})^{8}\)
\(P(X = 8)=1\times(\frac{1}{2})^{8}\)
\(P(X\gt4)=\frac{56 + 28+8 + 1}{256}=\frac{93}{256}\approx0.363\)

Step5: Calculate \(P(X = 8)\)

For \(D\), \(P(X = 8)=C(8,8)\times(\frac{1}{2})^{8}\times(\frac{1}{2})^{0}=\frac{1}{256}\approx0.004\)

Step6: Calculate \(P(X\leq5)\)

For \(E\), \(P(X\leq5)=1 - P(X\gt5)=1-(P(X = 6)+P(X = 7)+P(X = 8))\)
\(P(X = 6)=\frac{28}{256}\), \(P(X = 7)=\frac{8}{256}\), \(P(X = 8)=\frac{1}{256}\)
\(P(X\leq5)=1-\frac{28 + 8+1}{256}=\frac{256-(37)}{256}=\frac{219}{256}\approx0.855\)

Step7: Calculate \(P(X\geq7)\)

For \(F\), \(P(X\geq7)=P(X = 7)+P(X = 8)\)
\(P(X = 7)=\frac{8}{256}\), \(P(X = 8)=\frac{1}{256}\)
\(P(X\geq7)=\frac{8 + 1}{256}=\frac{9}{256}\approx0.035\)

Answer:

A. \(\frac{35}{128}\)
B. \(\frac{9}{256}\)
C. \(\frac{93}{256}\)
D. \(\frac{1}{256}\)
E. \(\frac{219}{256}\)
F. \(\frac{9}{256}\)