Question

A, B, C in order cut a pack of cards, replacing them after each cut, on condition that the first who cuts a spade shall win a prize. Find their respective chances.

Answer 1

Hint: As A, B and C cut spade in respective order then we know that if A cuts the spade in the first attempt, then there is no chance of winning of B and C. If A loses then B gets the chance and if B loses then C gets the chance. And if A, B and C all couldn’t cut spade in the first attempt then again A gets the chance and this will go continuously until one cuts spade. So, we have to use this data to find their chances. The formula for probability is also to be used.

Complete step by step solution:
Now, let us begin. We know that there are 52 cards in a pack of cards and there are a total of 13 spade cards in a pack of cards.
We also know that the probability of an event occurred is given by the formula
= \[\dfrac{\text{number of favorable outcomes}}{\text{total number of outcomes}}\]
So, we can say that the probability of winning the prize
= \[\dfrac{13}{52}=\dfrac{1}{4}\]
We know that the total probability is = 1
So the probability of losing the prize =
= \[1-\dfrac{1}{4}=\dfrac{3}{4}\]
Now the probability of winning the prize for A = P(A) = when A cuts the spade that is he cuts the spade in first attempt + he cuts the spade in his second attempt (i.e. all three are failed to cut the spade in their first attempt ) + he cuts the spade in his third attempt (i.e. all three are failed to cut the spade in their second attempt) + so on.
P(A) = (A win) + (A loss) \[\times \] (B loss) \[\times \] (C loss) \[\times \] (A win) + (A loss) \[\times \] (B loss) \[\times \] (C loss) \[\times \] (A loss) \[\times \] (B loss) \[\times \] (C loss) \[\times \] (A win) + …..
P(A) = \[\dfrac{1}{4}+\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{1}{4}+\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{1}{4}+.....\]
P(A) = \[\dfrac{1}{4}+{{\left( \dfrac{3}{4} \right)}^{3}}\dfrac{1}{4}+{{\left( \dfrac{3}{4} \right)}^{6}}\dfrac{1}{4}+.....\]
P(A) = \[\dfrac{1}{4}\left( 1+{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( \dfrac{3}{4} \right)}^{6}}+.... \right)...........(1)\]
Here \[\left( 1+{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( \dfrac{3}{4} \right)}^{6}}+.... \right)........(2)\] is a Geometric Progression whose first term a = 1 and the common ratio r = \[{{\left( \dfrac{3}{4} \right)}^{3}}\]
We know that the formula for summation of geometric progression is given by \[a\left( \dfrac{1-{{r}^{n}}}{1-r} \right)\] , where a is first term and r is common ratio.
Applying this formula in equation (2), we get
\[\left( 1+{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( \dfrac{3}{4} \right)}^{6}}+.... \right)\]= \[1\left( \dfrac{1-{{\left( {{\left( \dfrac{3}{4} \right)}^{3}} \right)}^{n}}}{1-{{\left( \dfrac{3}{4} \right)}^{3}}} \right)............(3)\]
But when \[n\to \infty \] , \[{{\left( {{\left( \dfrac{3}{4} \right)}^{3}} \right)}^{n}}\to 0\]
So equation (3) becomes
\[\dfrac{1}{\left( 1-{{\left( \dfrac{3}{4} \right)}^{3}} \right)}............(4)\]
\[\dfrac{1}{\left( 1-\dfrac{27}{64} \right)}=\dfrac{1}{\left( \dfrac{37}{64} \right)}=\dfrac{64}{37}\]
Now from equation (1), the probability of winning prize of A
= \[\dfrac{1}{4}\times \dfrac{64}{37}=\dfrac{16}{37}\]
So the chances of A to win the prize P(A) = \[\dfrac{16}{37}\]
Now the probability of winning the prize for B = when A lost to cut the spade and B cuts the spade in first attempt + B cuts the spade in second spade (i. e. All three failed to cut the spade in their first attempt and A failed to cut the spade in second attempt) + so on.
P(B) = (A loss) \[\times \] (B win) + (A loss) \[\times \] (B loss) \[\times \] (C loss) \[\times \] (A loss) \[\times \] (B win) + (A loss) \[\times \] (B loss) \[\times \] (C loss) \[\times \] (A loss) \[\times \] (B loss) \[\times \] (C loss) \[\times \] (A loss) \[\times \] (B win) + …..
P(B) = \[\dfrac{3}{4}\dfrac{1}{4}+\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{1}{4}+\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{1}{4}+.....\]
P(B) = \[\dfrac{3}{4}\dfrac{1}{4}+{{\left( \dfrac{3}{4} \right)}^{3}}\dfrac{3}{4}.\dfrac{1}{4}+{{\left( \dfrac{3}{4} \right)}^{6}}\dfrac{3}{4}.\dfrac{1}{4}+.....\]
P(B) = \[\dfrac{1}{4}.\dfrac{3}{4}\left( 1+{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( \dfrac{3}{4} \right)}^{6}}+.... \right)...........(5)\]
From equation (3), (4) and (5) we can say that
P(B) = \[\dfrac{1}{4}\times \dfrac{3}{4}\times \dfrac{64}{37}=\dfrac{12}{37}\]
So the chances of B to win the prize P(B) = \[\dfrac{12}{37}\]
Now the probability of winning the prize for C = when A and B lost to cut the spade and C cuts the spade in first attempt + C cuts the spade in second spade (i. e. All three failed to cut the spade in their first attempt and A and B failed to cut the spade in second attempt) + so on.
P(C) = (A loss) \[\times \] (B loss) \[\times \] (C win) + (A loss) \[\times \] (B loss) \[\times \] (C loss) \[\times \] (A loss) \[\times \] (B loss) \[\times \] (C win) + (A loss) \[\times \] (B loss) \[\times \] (C loss) \[\times \] (A loss) \[\times \] (B loss) \[\times \] (C loss) \[\times \] (A loss) \[\times \] (B loss) \[\times \] (C win) + …..
P(C) = \[\dfrac{3}{4}\dfrac{3}{4}\dfrac{1}{4}+\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{1}{4}+\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{3}{4}\dfrac{1}{4}+.....\]
P(C) = \[\dfrac{3}{4}\dfrac{3}{4}\dfrac{1}{4}+{{\left( \dfrac{3}{4} \right)}^{3}}\dfrac{3}{4}.\dfrac{3}{4}.\dfrac{1}{4}+{{\left( \dfrac{3}{4} \right)}^{6}}\dfrac{3}{4}.\dfrac{3}{4}.\dfrac{1}{4}+.....\]
P(C) = \[\dfrac{1}{4}.\dfrac{3}{4}.\dfrac{3}{4}\left( 1+{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( \dfrac{3}{4} \right)}^{6}}+.... \right)...........(6)\]
From equation (3), (4) and (6) we can say that
P(C) = \[\dfrac{1}{4}\times \dfrac{3}{4}\times \dfrac{3}{4}\times \dfrac{64}{37}=\dfrac{9}{37}\]
So the chances of C to win the prize P(C) = \[\dfrac{9}{37}\]
The probability for A to win = \[\dfrac{16}{37}\]
The probability for B to win = \[\dfrac{12}{37}\]
The probability for C to win = \[\dfrac{9}{37}\]

Note: Let us consider an alternate method for finding the probability of C. We know that total probability = 1 i.e. P(A) + P(B) + P(C) = 1
If we have the probability of A and B then we can easily find the probability of C that is
P(C) = 1 – P(A) – P(B)
P(C) = \[1-\dfrac{16}{37}-\dfrac{12}{37}\]
P(C) = \[\dfrac{9}{37}\]