Question

Three cards are drasn without replacement from the 12 cards that make up the 4s; 55 , and 6s from an ordinary deck of 52 plinying cards, Let X be the number of 6 s selecled and Y the number of 4 s. (a) Find the joint probability distribution of X and Y. (b) Find P(X,Y)∈A], where A is the region given by {(x,y){x+y≥1}, (a) Complote the joint probability distribution below. (Type integers or simpaifind tractions.)

Answer 1

(a) The joint probability distribution of X and Y is as follows:

Y=0 & Y=1 & Y=2

`X=0 & (6)/(33) & (15)/(33) & (12)/(33) \\`

`X=1 & (6)/(33) & (9)/(33) & (0)/(33) \\`

(b) To find
`\( P(X,Y) \in A \)`, where A is the region given by
`\( \{(x, y) \mid x + y \geq 1\} \)`, sum the probabilities for the corresponding values in the joint probability distribution where
`\( x + y \geq 1 \)`. The result is
`\( P(X,Y) = (15)/(33) + (12)/(33) + (9)/(33) = (36)/(33) \).`

Step-by-step explanation:

(a) To determine the joint probability distribution of X and Y, we need to consider the possible combinations of X and Y. Since three cards are drawn without replacement from the deck, the probabilities depend on the outcomes of previous draws.

The joint probability distribution is given by:

Y=0 & Y=1 & Y=2

`X=0 & (6)/(33) & (15)/(33) & (12)/(33) \\`

`X=1 & (6)/(33) & (9)/(33) & (0)/(33) \\`

`\end{array}\]`

(b) To find
`\( P(X,Y) \in A \)`, we sum the probabilities for the values in the joint probability distribution where
`\( x + y \geq 1 \)`. The resulting probability is
`\( P(X,Y) = (36)/(33) \).`