# The numbers 15, 16, 17, 18, and 19 are written on slips of paper, and 2 slips are drawn at random one at a time without replacement. Find the probability of each of the following: a) Both numbers are even. b) One of the numbers is even or greater than 17. c) The sum is 33 or the second number is 16.

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)