Question

A card is selected from a standard deck and replaced. This experiment is repeated 8 times. Find the probability of selecting exactly three clubs.

Answer 1

The probability of selecting exactly three clubs in eight draws, with replacement from a standard deck, is
`\( (3)/(64) \)`.

When a card is selected from a standard deck and replaced, the probability of selecting a club is
`\( (1)/(4) \)` since there are four suits in a deck, and one of them is clubs.

The probability of not selecting a club is
`\( 1 - (1)/(4) = (3)/(4) \)`.

Now, since each selection is independent (the card is replaced after each draw), we can use the binomial probability formula to find the probability of getting exactly three clubs in eight draws:

`\[ P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^(n-k) \]`

For this problem:

- n = 8 (number of draws),

- k = 3 (exactly three clubs),

-
`\( p = (1)/(4) \)` (probability of getting a club in a single draw),

-
`\( 1-p = (3)/(4) \)` (probability of not getting a club in a single draw).

The formula becomes:

`\[ P(X=3) = \binom{8}{3} \left((1)/(4)\right)^3 \left((3)/(4)\right)^5 \]`

Now, let's calculate this:

`\[ P(X=3) = \binom{8}{3} \cdot \left((1)/(4)\right)^3 \cdot \left((3)/(4)\right)^5 \]\[ P(X=3) = (8!)/(3!(8-3)!) \cdot \left((1)/(4)\right)^3 \cdot \left((3)/(4)\right)^5 \]\[ P(X=3) = (8 \cdot 7 \cdot 6)/(3 \cdot 2 \cdot 1) \cdot (1)/(64) \cdot (243)/(1024) \]\[ P(X=3) = 56 \cdot (1)/(64) \cdot (243)/(1024) \]\[ P(X=3) = (3)/(64) \]`

Therefore, the probability of selecting exactly three clubs in eight draws is
`\( (3)/(64) \)`.