MATH SOLVE

4 months ago

Q:
# An urn contains 4 white and 4 black balls. We randomly choose 4 balls. If 2 of them are white and 2 are black, we stop. If not, we replace the balls in the urn and again randomly select 4 balls. This continues until exactly 2 of the 4 chosen are white. What is the probability that we shall make exactly n selections?

Accepted Solution

A:

Answer:The probability that we shall make exactly n selections is [tex]P(X = n)=(\frac{17}{35})^{n-1}\frac{18}{35}[/tex].Step-by-step explanation:It is given that an urn contains 4 white and 4 black balls and we randomly choose 4 balls. If 2 of them are white and 2 are black, we stop.The total number of ways to select exactly 2 white and 2 black balls.[tex]^4C_2\times ^4C_2=\frac{4!}{2!(4-2)!}\times \frac{4!}{2!(4-2)!}=6\times 6=36[/tex]The total number of ways to select 4 balls from 8 balls is[tex]^8C_4=\frac{8!}{4!(8-4)!}=\frac{8\times 7\times 6\times 5\times 4!}{4\times 3\times 2\times 1\times !4!}=70[/tex]The probability of selecting exactly 2 white and 2 black balls is[tex]p=\frac{36}{70}=\frac{18}{35}[/tex]The probability of not selecting exactly 2 white and 2 black balls is[tex]q=1-p=1-\frac{18}{35}=\frac{17}{35}[/tex]If we not get exactly 2 white and 2 black balls, then we replace the balls in the urn and again randomly select 4 balls. The probability that we shall make exactly n selections is [tex]P(X = n)=(q)^{n-1}p[/tex][tex]P(X = n)=(\frac{17}{35})^{n-1}\frac{18}{35}[/tex]Therefore the probability that we shall make exactly n selections is [tex]P(X = n)=(\frac{17}{35})^{n-1}\frac{18}{35}[/tex].