A new surgery is successful 75% of the time. if the results of 10 such surgeries are randomly sampled, what is the probability that fewer than 9 of them are successful

ANSWERS

2016-04-29 14:34:23

[latex] ext{Binomial Distribution: } P(X = r) = inom{n}{r}(p)^{r}(1 - p)^{n - r}[/latex] [latex]P(X < 9) = 1 - P(X = 9) - P(x = 10)[/latex] [latex]P(X = 9) = inom{10}{9}(0.75)^{9}(0.25)[/latex] [latex]P(X = 10) = inom{10}{10}(0.75)^{10}[/latex] [latex]P(X < 9) = 1 - inom{10}{9}(0.75)^{9}(0.25) - inom{10}{10}(0.75)^{10}[/latex] [latex] approx 0.755974769...[/latex]

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