# Probability

• Dec 3rd 2009, 12:23 PM
BabyMilo
Probability
A and B take it in turns, starting with A, to take a card, without replacement, from a pack of 10 cards containing the numbers 1,2,2,3,3,3,4,4,4,4.

The first player to select a "4" card is the winner.

Find the probability that B wins the game.
• Dec 3rd 2009, 02:19 PM
Soroban
Hello, BabyMilo!

Quote:

Starting with $A$, players $A$ and $B$ take turns taking a card, without replacement,
from a pack of 10 cards marked: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4.

The first player to select a "4" card is the winner.

Find the probability that $B$ wins the game.

There are: 4 Fours and 6 Others.

A tree diagram makes the problem clear.

There are three scenarios in which $B$ wins.

[1] $A$ draws an Other, then $B$ draws a Four.
. . .This probability is: . $\frac{6}{10}\cdot\frac{4}{9} \:=\:\frac{4}{15}$

[2] $A$ draws an Other, $B$ draws an Other,
. . . $A$ draws an Other, then $B$ draws a Four.
. . .This probability is: . $\frac{6}{10}\cdot\frac{5}{9}\cdot\frac{4}{8}\cdot\ frac{4}{7} \:=\:\frac{2}{21}$

[3] $A$ draws an Other, $B$ draws an Other,
. . . $A$ draws an Other, $B$ draws an Other,
. . . $A$ draws an Other, then $B$ draws a Four.
. . .This probability is: . $\frac{6}{10}\cdot\frac{5}{9}\cdot\frac{4}{8}\cdot\ frac{3}{7}\cdot\frac{2}{6}\cdot\frac{4}{5} \:=\:\frac{2}{105}$

Therefore: . $P(B\text{ wins}) \;=\;\frac{4}{15} + \frac{2}{21} + \frac{2}{105} \;=\;\frac{8}{21}$

But check my work . . . please!
.