Alice, Benjamin, and Carol each try independently to win a c

BTGmoderatorDC wrote:Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?

A. 3/140
B. 1/28
C. 3/56
D. 3/35
E. 7/40
Source: Veritas Prep

$$? = P\left( {A \cap B \cap \overline C } \right) + P\left( {A \cap C \cap \overline B } \right) + P\left( {B \cap C \cap \overline A } \right)$$
$$\left( * \right)\,\,\left\{ \matrix{
\,P\left( {A \cap B \cap \overline C } \right) = {1 \over 5} \cdot {3 \over 8} \cdot {5 \over 7} = {{15} \over {5 \cdot 8 \cdot 7}} \hfill \cr
\,P\left( {A \cap C \cap \overline B } \right) = {1 \over 5} \cdot {2 \over 7} \cdot {5 \over 8} = {{10} \over {5 \cdot 7 \cdot 8}} \hfill \cr
\,P\left( {B \cap C \cap \overline A } \right) = {3 \over 8} \cdot {2 \over 7} \cdot {4 \over 5} = {{24} \over {8 \cdot 7 \cdot 5}} \hfill \cr} \right.$$
$$? = {{49} \over {5 \cdot 7 \cdot 8}} = {7 \over {40}}$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

The probability that exactly two of the three players will win = probability of (A will win, B will win, C will lose + A will win, B will lose, C will win + A will lose, B will win, C will win)

=> (1/5 * 3/8 * (1- 2/7)) + (1/5 * (1 - 3/8) * 2/7) + ((1 - 1/5) * 3/8 * 2/7)
=> 15/280 + 10/280 + 24/280
=> 49/280
=> 7/40