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

you can not drow on this

step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.