Question

A bag contains some white and some black balls, all combinations being equally likely. The total number of balls in the bag is $12$. Four balls are drawn at random without replacement.

find the following:

-Probability that all the balls are black is equal to

-If the bag contains 10 black and 2 white balls, then the probability that all four balls are black is equal to

-If all the four balls are black , then the probability that the bag contains 10 black balls is equal to

Answer 1

If the bag contains 10 black and 2 white balls, the probability that all four balls are black is equal to $\frac{{}^{10}{C}_4}{{}^{12}{C}_4}=\frac{8\times 7}{12\times 11}=\frac{14}{33}$

Let E${}_i$: an event that there are i black balls, (12-i) white balls, where i= 0, 1, 2, 3, ... 12 i.e., 13 outcomes with each combination equally likely

And A: an event that all the four balls drawn are black.

Then, $P\left(E_i\right)=\frac{1}{13}$

and, $P\left(A/E_i\right)=0$ when i = 0, 1, 2, 3 as the number of black balls drawn are 4. and $P\left(A/E_i\right)=\frac{{}^i{C}_4}{{}^9{C}_4}$ when i = 4, 5, 6, ... 12 = 9 outcomes.

Hence $P\left(A\right)={\sum }_{i=4}^{12}\left(P\right(E_i)\times P(A/E_i\left)\right)⟹P\left(A\right)=\frac{1}{13}(\frac{{}^4{C}_4}{{}^9{C}_4}+\frac{{}^5{C}_4}{{}^9{C}_4}+\frac{{}^6{C}_4}{{}^9{C}_4}+\frac{{}^7{C}_4}{{}^9{C}_4}+\frac{{}^8{C}_4}{{}^9{C}_4}+\frac{{}^9{C}_4}{{}^9{C}_4}+\frac{{}^{10}{C}_4}{{}^9{C}_4}+\frac{{}^{11}{C}_4}{{}^9{C}_4}+\frac{{}^{12}{C}_4}{{}^9{C}_4})$

${}^4{C}_4$ can be written as ${}^5{C}_5$

and we know ${}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$

Hence ${}^4{C}_4{+}^5{C}_4{+}^6{C}_4{+}^7{C}_4+...{+}^{12}{C}_4{=}^5{C}_5{+}^5{C}_4{+}^6{C}_4{+}^7{C}_4{+}^{.}..{+}^{12}{C}_4{=}^6{C}_5{+}^6{C}_4{+}^7{C}_4+...{+}^{12}{C}_4{=}^7{C}_5{+}^7{C}_4+...{+}^{12}{C}_4⟹{=}^{12}{C}_5{+}^{12}{C}_4{=}^{13}{C}_4$

Therefore $P\left(A\right)=\frac{1}{13}\times \frac{1}{{}^9{C}_4}{\times }^{13}{C}_5=\frac{13\times 12\times 11\times 10\times 9}{13\times 9\times 8\times 7\times 6\times 5}=\frac{11}{14}$

is the probability that all the balls are black.

Now here we need to find Probability of 10 black balls when event A has already occured, i.e.,$P\left(E_{10}/A\right)=\frac{P(E_{10}\cap A)}{P\left(A\right)}$

$⟹P\left(E_{10}/A\right)=\frac{P\left(E_{10}\right)\times P\left(A/E_{10}\right)}{P\left(A\right)}=\frac{\frac{1}{13}\times \frac{{}^{10}{C}_4}{{}^9{C}_4}}{\frac{11}{14}}=\frac{1\times 14\times 10!\times 4!\times (9-4)!}{13\times 11\times 4!\times (10-4)!\times 9!}=\frac{14\times 10}{13\times 11\times 6}=\frac{70}{429}$

is the required probability that the bag contains 10 black balls.