Answer to Question #213196 in Statistics and Probability for mel

Question

Question 1.3 [4]

A set of eight cards contains one joker. Aviwe and Siphumlile are two players that are learning probability theory by playing card games with these 8 cards. Aviwe chooses 5 cards at random, Siphumlile takes the remaining three cards. What is the probability that Aviwe has the joker? Aviwe discards 4 cards and Siphumlile discards two cards. It is know that the joker has not been discarded, what is the probability that Aviwe has the joker? Explain your answer in detail.

Answer 1

There are 7 cards and 1 joker. Hence a total of 8 cards.

First, Aviwe chose 5 cards at random .

The total combinations that Aviwe chooses of the 5 cards is $8C_5=56$

The total combinations that Siphumlile chooses of the 3 cards is $8C_3=56$

Probability that Aviwe has a joker is calculated as:

$\:\frac{7C_4\times 1C_1}{8C_5}\times \frac{7C_3\times 1C_0}{8C_3}=\frac{35\times 1}{56}\times \frac{35\times 1}{56}=\frac{25}{64}$

Aviwe discards 4 cards and Siphumlile discards 2 cards. Jockler has not been discarded.

Now Aviwe has 1 card and Siphumlile has 1 card.

$P\left(Aviwe\:has\:a\:Joker\right)=\frac{4C_0\times 1C_1}{5C_1}\times \frac{3C_1}{3C_1}=\frac{1}{5}\times 1=\frac{1}{5}$

Therefore, the probability that Aviwe has a Joker is $\frac{1}{5}$ .