We are given the set {15,16,17,18,19} and we will draw 2 numbers out, sequentially without replacement:

**a) What is the probability that both numbers are even?**

One way is to use the multiplication principle: take the probability that the first even occurs times the probability that the second even occurs.

The probability that the first number drawn is even is 2/5 or 0.4. The number of events in the event space is 2 (the number of even numbers), while the number of things in the sample space is 5 (the total number of possible events).

Now we use conditional probability. Assuming the first number is even, the probability that the second number is even is 1/4 or 0.25. (1 even number left and 4 numbers to choose from.)

So, the probability that both are even is (2/5)(1/4)=1/10.

(ii) We can directly find P(E,E) by `P(E)*P(E|E)=.4*.25=.1`

(iii) We could draw a tree diagram. There are 20 possible pairs, only two of which have both even numbers. (16,18) and (18,16)

**(b) Find the probability that one of the numbers is even or greater than 17.**

(i) We can use the complement. If the probability is P we will denote the complement as P'. (There is no standard notation; some texts use C(P), others `bar(P)` , etc...) P=1-P'. Here the complement is 1/10. (There are two pairs whose members are neither even nor greater than 17: (15,17) and (17,15) and there are 20 total pairings.)

Thus the probability we want is 1-1/10=9/10.

(ii) P(E)+P(>17)-P(e and >17)

P(exactly one even)=`2* [(2C1)(3C1)/20]=12/20`

P(#>17)=`2*[2C2]\20+2*[(2C1)(3C1)]/20=14/20`

P(even and >17)=`2*[(1C1)(4C1)]/20=8/20`

Where 2C1 indicates the number of combinations of 2 items taken one at a time.

So 12/20+14/20-8/20=18/20=9/10 as before.

Use a tree.

**(c) The sum is 33 or the second number is 16.**

Sum equalling 33 implies 16+17 or 18+15 which happens 4/20 times.

Second number being 16 happens 4/20 times.

17+16 is counted twice (2nd number is 16 and sum=33), so

4/20+4/20-1/20= 7/20.

(16,17),(15,18),(18,15),(15,16),(17,16),(18,16),(19,16)