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.
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.
Hello, BabyMilo!
There are: 4 Fours and 6 Others.Starting with $\displaystyle A$, players $\displaystyle A$ and $\displaystyle B$ 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 $\displaystyle B$ wins the game.
A tree diagram makes the problem clear.
There are three scenarios in which $\displaystyle B$ wins.
[1] $\displaystyle A$ draws an Other, then $\displaystyle B$ draws a Four.
. . .This probability is: .$\displaystyle \frac{6}{10}\cdot\frac{4}{9} \:=\:\frac{4}{15}$
[2] $\displaystyle A$ draws an Other, $\displaystyle B$ draws an Other,
. . .$\displaystyle A$ draws an Other, then $\displaystyle B$ draws a Four.
. . .This probability is: .$\displaystyle \frac{6}{10}\cdot\frac{5}{9}\cdot\frac{4}{8}\cdot\ frac{4}{7} \:=\:\frac{2}{21}$
[3] $\displaystyle A$ draws an Other, $\displaystyle B$ draws an Other,
. . .$\displaystyle A$ draws an Other, $\displaystyle B$ draws an Other,
. . .$\displaystyle A$ draws an Other, then $\displaystyle B$ draws a Four.
. . .This probability is: .$\displaystyle \frac{6}{10}\cdot\frac{5}{9}\cdot\frac{4}{8}\cdot\ frac{3}{7}\cdot\frac{2}{6}\cdot\frac{4}{5} \:=\:\frac{2}{105}$
Therefore: .$\displaystyle P(B\text{ wins}) \;=\;\frac{4}{15} + \frac{2}{21} + \frac{2}{105} \;=\;\frac{8}{21}$
But check my work . . . please!
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