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Sagot :
Answer:
a) {3,5}{3,10}{5,10}
b) [tex]P(A)=\frac{1}{3}[/tex]
c) [tex]P(B)=\frac{2}{3}[/tex]
d) [tex]P(C)=\frac{1}{3}[/tex]
e) [tex]P(A and C)=0[/tex]
f) [tex]P(A or B)=1[/tex]
g) [tex]P(B and C)=\frac{1}{3}[/tex]
h) [tex]P(A or C)=\frac{2}{3}[/tex]
i) [tex]P(C given B)=\frac{1}{2}[/tex]
j) [tex]P(C given A)=0[/tex]
k) [tex]P(not B)=\frac{1}{3}[/tex]
l) [tex]P(not C)=\frac{2}{3}[/tex]
Yes, events A and B are mutually exclusive. Because the results can either be even or odd, not both. No, events B and C are not mutually exclusive because the result can be both, odd and prime.
Step-by-step explanation:
a)
In order to solve part a of the problem, we need to find the possible outcomes, in this case, the possible outcomes are:
{3,5}{3,10} and {5,10}
We could think of the oppsite order, for example {5,3}{10,3}{10,5} but these are basically the same as the previous outcomes, so we will just take three outcomes in our sample space. We can think of it as drawing the two chips at the same time.
b)
Now the probability of the sum of the chips to be even. There is only one outcome where the sum of the chips is even, {3,5} since 3+5=8 the other outcomes will give us an odd number, so:
[tex]P=\frac{#desired}{#possible}[/tex]
[tex]P(A)=\frac{1}{3}[/tex]
c) For the probability of the sum of the chips to be odd, there are two outcomes where the sum of the chips is odd, {3,10} since 3+10=13 and {5,10} since 5+10=15 the other outcomes will give us an even number, so:
[tex]P(B)=\frac{2}{3}[/tex]
d) The probability of the sum of the chips is prime. There is only one outcome where the sum of the chips is prime, {3,10} since 3+10=13 the other outcomes will give us non prime results, so:
[tex]P(C)=\frac{1}{3}[/tex]
e) The probability of the sum of the chips to be even and prime. There are no results where we can get an even and prime number, since the only even and prime number there is is number 2 and no outcome will give us that number, so:
P(A and C)=0
f) The probability of the sum of the chips is even or odd. We can either get even or odd results, so no matter what outcome we get, we will get an odd or even result so:
[tex]P(A or B)=1[/tex]
g) The probability of the sum of the chips is odd and prime. There is only one outcome where the sum of the chips is odd and prime, {3,10} since 3+10=13 the other outcomes will give us non prime results, so:
[tex]P(B and C)=\frac{1}{3}[/tex]
h) The probability of the sum of the chips is even or prime. There are two outcomes where the sum of the chips is even or prime, {3,10} since 3+10=13 and {3,5} since 3+5=8 so:
[tex]P(A or C)=\frac{2}{3}[/tex]
i) The probability of the sum of the chips is prime given that the sum of the chips is odd. There are two possible results where the sum of the chips is odd {3,10} and {5,10} and only one of those results is even, {3,10}, so
[tex]P(C given B)=\frac{1}{2}[/tex]
j) The probability of the sum of the chips is prime given that the sum of the chips is even. There is only one possible even result: {3,5} but that result isn't prime, so
[tex]P(C given A)=0[/tex]
k) The probability of the sum of the chips is not odd. There is only one outcome where the sum of the chips is not odd (even), {3,5} so:
[tex]P(not B)=\frac{1}{3}[/tex]
l) The probability of the sum of the chips is not prime. There are two outcomes where the sum of the chips is not prime, {3,5} and {5,10} so:
[tex]P(not C)=\frac{2}{3}[/tex]
Are events A and B mutually exclusive?
Yes, events A and B are mutually exclusive.
Why or why not?
Because the results can either be even or odd, not both.
Are events B and C mutually exclusive?
No, events B and C are not mutually exclusive.
Why or Why not?
Because the result can be both, odd and prime.
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