Question

three balls are drawn in succession without replacement from a jar containing 8 white balls and 5 black balls. Let B be the random variable representing number of black balls

Answer 1

Since it is not stated which parameter has to be calculated, consider possible values that the random variable $B$ may take. Since the number of balls are $3$, the random variable $B$ may take the following values: $0,1,2,3.$ The total number of balls is $8+5=13$. There are $C_{13}^3=\frac{13!}{10!3!}=\frac{11\cdot12\cdot13}{3!}=286$ different ways to choose $3$ balls from $13$. There are $C_8^{3}=\frac{8!}{5!3!}=\frac{6\cdot7\cdot8}{6}=56$ different ways to choose $3$ white balls from $8$. We receive that: $P(B=0)=\frac{C_8^3}{C_{13}^3}=\frac{56}{286}\approx0.196$. (it is rounded to $3$ decimal places) There are $C_8^2=\frac{{8!}}{6!2!}=28$ ways to choose $2$ white balls from $8$. There are $C_5^1=5$ different ways to choose $1$ black ball from $5$. Using the multiplication principle of combinatorics, we get: $P(B=1)=\frac{C_8^2C_5^1}{C_{13}^3}=\frac{28\cdot 5}{286}\approx0490.$ We continue further without explanations and receive: $P(B=2)=\frac{C_8^1C_5^2}{C_{13}^3}=\frac{8\cdot10}{286}\approx0.280$, $P(B=3)=\frac{C_5^3}{C_{13}^3}\approx0.035$.

Answer: $P(B=0)\approx0.196$, $P(B=1)\approx0.490$, $P(B=2)\approx0.280$, $P(B=3)\approx0.035$.(all values are rounded to $3$ decimal places)