Probabilitty Questions. HELLPPP!!?

Question

Two distinct #s are randomly chosen from the set {1,2,3,4,5,6}. Find the probability of each of the following events:

1. Both numbers are even

2. Both numbers are prime

3. The sum of the two numbers is even

4. The product of the two numbers is odd

Answer 1

(1) Even numbers: 2, 4, 6 = 3 choices
odds = 3/6 * 2/5 = 6/30 = 1/5

(2) Prime numbers: 1, 2, 3, 5 = 4 choices
odds = 4/6 * 3/5 = 12/30 = 2/5

(3) Even sum: 1+3, 1+5, 2+4, 2+6, 3+5, 4+6
odds = 6/30 = 1/5

(4) Odd product: 1*3, 1*5, 3*5
odds = 3/30 = 1/10

Answer 2

Assumes no replacement

2,4,6 => 3 #' out of 6 = P=.5 of picking one
having picked one now there are 2 evens out of 5 numbers left => 2/5 = P = .4
Over all Prob => P1*P2 = .5*.4 = .2 or 20%

2.) use same methodology as above accept prime #'s are 1,3,5 => same quantity therefore same answer

3.) The first number pick can be odd or even and there is a .5 Prob of it being one or the other.

if an even number is picked then another even must be pick for the sum to be even
P1=.5 P2=2/5 =.4 => Pa=P1*P2=.2
if an odd number is picked then an even must be picked for the sum to be even
P1=.5 p2=3/5 =.6 =>Pb=P1*P2 = .3
Since either of the two above will give you the desired outcome their probabilities are added together
P=Pa+Pb=.2+.3=.5 or 50%

4.) use the same logic as in #3 except notice that and odd*even = even and an even*odd = even and even*evan=even
first draw odd second even => .5*.6 =.3
first draw even second odd =>.5*.6 =.3
first draw even second even => .5*.4 =.2

P=.3+.3+.2 = .8 or 80%

Answer 3

1. P(Both #'s even) = p(first number even) * p(second number even) = 3/6 * 2/5 = 6/30 = 1/5

2. P(Both #'s prime) = p(first number prime) * p(second number prime) = 4/6 * 3/5 = 2/5

3. P(sum is even) = since both numbers have to be odd or even then so = p(both numbers even) + p(both number odd) = 1/5 + 1/5 = 2/5

4. P(product is odd) = then both numbers have to be odd = p(first number odd) * p(second number odd) = 3/6 * 2/5 = 1/5

i believe that my answers are correct.

Answer 4

uhh it depends on whether or not you can reuse the numbers for both cases.

both numbers are even, can they both be 4 or does one have to be 2 or 6?

Answer 5

1. 3/6 = 1/3
2. 1/3
3. 3/6
4. 3

Answer 6

even numbers:- 2,4,6
P(e) = total even/total numbers
P(e) = 3/6
P(e) = 1/2

Prime numbers:- 1,2,3,5
P(p) = 4/6
P(p) = 2/3

Sum Even:- (1,3), (2,4), (2,6), (4,6), (3,5), (1,5)
P(se) = 6/6
P(se) = 1

Product odd:-(1,3), (3,5), (1,5)
P(po) = 3/6
P(po) = 1/2

P(e) = 1/2
P(p) = 2/3
P(se) = 1
P(po) = 1/2