Suppose you first randomly sample one card from a deck of 52. Then, without putting the card back in the deck you sample a second and then (again without replacing cards) a third. Given this sampling procedure, what is the probability that at least one ace will be in the three sampled cards

P(A) = 1201/5525 or 0.2174

The probability that at least one ace will be in the three sampled cards is 1201/5525 or 0.2174

Step-by-step explanation:

Let P(A) represent the probability that at least one ace will be in the three sampled cards.

P(A) = 1 - P(A') 1

Where;

P(A') is the probability that none of the three cards will be ace.

Number of ace in a standard 52 cards = 4

Number of non-ace in a standard 52 cards = 52-4 =48

P(A') = 48/52 × 47/51 × 46/50

P(A') = 4324/5525 = 0.7826

Substituting, the value into equation 1;

P(A) = 1 - P(A')

P(A) = 1 - 4324/5525

P(A) = 1201/5525 or 0.2174

The probability that at least one ace will be in the three sampled cards is 1201/5525 or 0.2174

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + * x

4500-875 =   * x

3625= * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x =  

x= $2416.6