Initially, a standard deck of cards has 52 cards.
There are 13 cards of each suit: hearts, diamonds, clubs, and spades.
Hearts and diamonds are red, while clubs and spades are black.
So there are 26 red cards and 26 black cards.
We are given that 2 cards of hearts are missing and 4 cards of spades are missing.
The total number of missing cards is $2 + 4 = 6$.
The number of cards remaining is $52 - 6 = 46$.
Originally, there were 13 spades, but 4 are missing.
So the number of spades remaining is $13 - 4 = 9$.
Originally, there were 13 clubs, and no clubs are missing.
So the number of clubs is 13. The total number of black cards remaining is $9 + 13 = 22$.
The probability of getting a black card from the remaining pack is the number of black cards remaining divided by the total number of cards remaining: $$ P(\text{black card}) = \frac{\text{Number of black cards remaining}}{\text{Total number of cards remaining}} = \frac{22}{46}$$
If the probability distribution is given by:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| P(x) | 0 | k | 2k | 2k | 3k | k² | 2k² | 7k² + k |
Then find: \( P(3 < x \leq 6) \)
If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :