Answer:
a. 0.349
b. 0.651
c. 0.93
Step-by-step explanation:
According to the Question,
Thus, p=0.10 , n=10 & Â [tex]\left[\begin{array}{ccc}n\\x\\\end{array}\right] = \frac{n!}{x!(n-x)!}[/tex] Â
Now, Use Binomial distribution = [tex]\left[\begin{array}{ccc}n\\x\\\end{array}\right]*(p)^{x} *{(1-p)}^{n-x}[/tex]
a. p(x=0) ⇒  [tex]\left[\begin{array}{ccc}10\\0\\\end{array}\right]*(0.1)^{0} *{(0.9)}^{10}[/tex] = 0.349 Â
Thus, The Probability that No one has done a one-time fling = 0.349
b. p(x≥1) = 1 - p(x=0)  ⇒  1 - 0.349 = 0.651
Thus, The Probability that at least one person has done a one-time fling = 0.651
c. p(x≤2) = p(x=0) + p(x=1) + p(x=2)
p(x≤2) = [tex]\left[\begin{array}{ccc}10\\0\\\end{array}\right]*(0.1)^{0} *{(0.9)}^{10}[/tex] + [tex]\left[\begin{array}{ccc}10\\1\\\end{array}\right]*(0.1)^{1} *{(0.9)}^{9}[/tex] + [tex]\left[\begin{array}{ccc}10\\2\\\end{array}\right]*(0.1)^{2} *{(0.9)}^{8}[/tex]
on Solving, we get
p(x≤2) = 0.349 + 0.347 + 0.194
p(x≤2) = 0.93
Thus, The Probability that No More Than two person have done a one-time fling = 0.93