# Calculate conditional probability on drawing cards

In a game of drawing cards, $4$ cards will be drawn from the deck randomly. In the deck of cards, $0.4$ of them are black, $0.6$ of them are white. Among the black cards, $20\%$ of them are worth $1$ point, $50\%$ of them are worth $0$, and $30\%$ of them are worth $-1$ point. Among the white cards, $20\%$ of them are worth $1$ point, $25\%$ are worth $0$, and $55\%$ of them are worth $-1$ point.

Given that a player has drawn $3$ black and $1$ white, what is the probability that there are exactly two $+1$ point cards and one $+0$ points cards? Answer provided is $0.078$.

What I have attempted is that I divide into 4 cases. The player can draw (+1,+1,0), (+1,+1,-1), (+1,0,-1) and (+1,0,-1), assuming all cards are distinct, from the black cards. And I compute the following

$$\frac{\{(0.4^3)(0.6)\}\{(0.2)(0.2)(0.5)(0.55)+(0.2)(0.2)(0.3)(0.25)+2(0.2)(0.5)(0.3)(0.2)\}}{(0.4)^3(0.6)}$$

What did I miss out? Or is my thinking totally wrong? Thank you in advance.

• You need the large of the deck... this is an infinite deck? They are 10 or more cards in the deck? The things change drastically depending of the large of the deck. – Masacroso Dec 10 '15 at 12:05
• Total number of cards is not stated in the question. Just know 0.4 are black, and 0.6 are white. – Jacky Dec 10 '15 at 12:07
• So we must assume the deck is infinite, i.e. there is replacement. In other case exact probabilities are unknown. – Masacroso Dec 10 '15 at 12:10

I assume the deck is infinite so there are replacement and the probabilities stay constant. In other case the probabilities change in some degree depending of the number of cards on the deck.

We have 4 cards, 3 black and 1 white, and we want to count the probability for the combinations with two +1 cards, one 0 card and one -1 card. We have 3 cards with the same probability distribution so it will be more counting from any different valour of the white card:

1) If the white card is $+1$ then the black cards must be $(+1,0,-1)$ i.e. $P(W=+1)=0.2^2\cdot 0.5\cdot 0.3\cdot \color{red}{3!}=0.036$

2) If the white card is $0$ then black cards must be $(-1,+1,+1)$ i.e. $P(W=0)=0.25\cdot0.2^2\cdot0.3\cdot\color{red}{3}=0.009$

3) If the white card is $-1$ then w have on blacks $(+1,+1,0)$ so $P(W=-1)=0.55\cdot0.2^2\cdot0.5\cdot\color{red}{3}=0.033$

So the total is $0.036+0.009+0.033=0.078$. Two keys here:

a) You dont need to take into account the general probabilities of $0.6$ and $0.4$ because the cards are already taken.

b) For every case you need to take into account the different possible combinations, that are marked on red. The blacks cards have more than one way to combine to form the desired group, in the first group all are different so we have $3!$ ways that the black cards can form the $(-1,0,+1)$ group. In the other 2 cases they are $3!/2!=3$ ways that the black cards can be arranged, dividing by $2!$ is needed because if not we are counting twice some combinations because we have the double ways to form the same group because we have two identical cards.