Question

An urn contains 12 balls, of which 4 are white. Three players A, B, and C successively draw from the urn, A first, then B, then C, then A, and so on. The winner is the

first one to draw a white ball. Find the probability of winning

for each player if

(a) each ball is replaced after it is drawn;

(b) the balls that are withdrawn are not replaced.

Short answer

(a). A, B, and C can win chances $\frac{9}{19},\frac{6}{19},\frac{4}{19}$singly if the balls are drawn with relief.

(b).$\text{A, B, and}C\text{can win probabilities}\frac{77}{165},\frac{68}{165},\frac{20}{165}\text{respectively if the balls are drawn without replacement.}$

Step-by-step solution

Explanation (Part a)

Given: An casket contains 4 balls out of which 4 are white. A, B, C draw one ball from the casket in race. The earliest one to pull a white ball wins.
Formula used :

Probability of an event$=\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}}$

Sum of chances of all possible outcomes is 1

Sum of an horizonless GP with common rate $r<1$and first term a is$\frac{a}{1-r}$

If the balls are drawn with relief , A can win in the first turn with probabilitylocalid="1647404163506" $\frac{4}{12}=\frac{1}{3}$

If A doesn't win in the first turn with probabilitylocalid="1647404173666" $1-\frac{1}{3}$. Also B and C should both withdraw non-white balls with probabilitylocalid="1647404187995" $\frac{2}{3}$. Also A can win in the alternate turn with the probability$\frac{1}{3}·{\left[\frac{2}{3}\right]}^3$

Also, the chances of A winning in the$\frac{1}{3}\sum _{i=1}^{\infty }{\left[\frac{2}{3}\right]}^{3\left(i-1\right)}=\frac{9}{19}$

Hence, probability of A winning is $\frac{1}{3}\sum _{i=1}^{\infty }{\left[\frac{2}{3}\right]}^{3(i-1)}=\frac{9}{19}$

Also , B can win in the first turn with the probabilitylocalid="1647404208168" $\frac{2}{3}·\frac{1}{3}$if A loses in the first turn.

Calculation (Part a)

B can win in the alternate turn if A, B, and C lose in the first turn each with probability $\frac{2}{3}$and A loses in the alternate turn again with the same probability. Hence, B can win in the alternate turn with the probability$\frac{1}{3}·{\left[\frac{2}{3}\right]}^4$

B can win with probability$\frac{1}{3}·\frac{2}{3}\sum _{i=1}^{\infty }{\left[\frac{2}{3}\right]}^{3(i-1)}=\frac{2}{9}\frac{1}{19}=\frac{6}{19}$.

C can with probability$1-\frac{9}{19}-\frac{6}{19}=\frac{4}{19}$

A, B, and C can win chances$\frac{77}{165},\frac{68}{165},\frac{20}{165}$singly if the balls are drawn without relief.

Explanation (Part b)

Given: An charnel contains 4 balls out of which 4 are white. A, B, C draw one ball from the charnel in race. The earliest one to drag a white ball wins.

Formula used:

Probability of an event$=\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}}$

Sum of chances of all possible issues is$1.$

Sum of an bottomless GP with common rate$r<1$and first term a is$\frac{a}{1-r}$.

If the balls are not replaced,

Still, can win in the first turn with the probability$\frac{1}{3}$

If A does not win in the first turn,

Still, B and C must also draw non-white balls. After A has drawn a non-white ball in the first turn, B has 7 non-white balls to draw from while C has 6 non-white balls to draw from. Hence, A can win in the alternate turn with the probability$\frac{8}{12}·\frac{7}{11}·\frac{6}{10}·\frac{4}{9}$

Calculation (Part b)

A has a last chance of winning in the third turn if A, B, C draw non-white balls in first two turns. Probability of A winning in the third turn$\frac{8\times 7\times 6\times 5\times 4\times 3}{12\times 11\times 10\times 9\times 8\times 7}.\frac{4}{6}$

Hence, probability of A winning is sum of chances of winning in first, alternate, and third term$\frac{7}{15}$

B can win in the first turn if A draws a non-white ball. The probability of B winning is$\frac{11}{12}.\frac{4}{11}$

B can win in the alternate turn if all three players draw non-white balls with probability $\frac{8\times 7\times 6\times 5}{12\times 11\times 10\times 9}·\frac{4}{8}$.

B can win in the third turn if all three players draw non-white balls with probability $\frac{8\times 7\times 6\times 5\times 4\times 3\times 2}{12\times 11\times 10\times 9\times 8\times 7\times 6}·\frac{4}{5}$.

Hence, probability of B winning is sum of all chances in each turn$=\frac{68}{165}.$

Probability of C winning$=1-\frac{7}{15}-\frac{68}{165}=\frac{20}{165}.$