Question

The probability that a male professional golfer makes a hole-in-one is 1/2780. Suppose 36 professional male golfers play the sixth hole during a round of golf. Let the random variable X be the number of golfers in the group of 36 who make a hole-in-one. Calculate the probability that exactly four of the 36 golfers make a hole-in-one on the sixth hole – as actually happened during the 1989 U.S. Open

Answer 1

Answer:

The right solution  is "$9.7\times 10^{-10}$".

Step-by-step explanation:

According to the question,

The probability that male professional golfer makes hole in one will be:

$P=\frac{1}{2780}$

Number of players,

n = 36

and,

$q=1-P=\frac{2779}{2780}$

By using the Binomial theorem, we get

⇒ $P(x=r) = \binom{n}{r} p^r q^{n-r}$

Bu substituting the values, we get

                   $=\binom{36}{4} (\frac{1}{2780} )^4 (\frac{2779}{2780} )^{32}$

                   $=9.74929\times 10^{-10}$

or,

                   $=9.7\times 10^{-10}$