A box contains 5 tickets numbered 1,2,3,4, and 5. Two tickets are drawn at random from the box. Find the chance that the numbers on the two tickets differ by two or more if the draws are made: a) with replacement; b) without replacement. Repeat the problem with n tickets numbered 1,2, ... ,n.

(a) The probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) The probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

Step-by-step explanation:

The tickets are drawn such that the difference between the two numbers is at least 2.

That is, 1st number - 2nd number ≥ 2.

The sample space such that this condition is satisfied is:

S = {(1, 3), (1, 4), (1, 5), (2, 4), (2, 5) and (3, 5)} = 6 possible pairs.

But there is also case where the numbers can be drawn in reverse order, i.e we can draw (3, 1) instead of (1, 3).

This makes the total number of possible pairs as, 6 × 2 = 12 pairs.

(a) With Replacement

In case the tickets are selected with replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) Without Replacement:

In case the tickets are selected without replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

∠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.