Question

a 6 - sided die is rolled once. find

1. p(roll a 5)
2. p(roll a 1 or 6)
3. p(odd #)
4. p(multiple of 3)
5. p(not a 4)

a card is drawn from a standard 52 - card deck. find:

1. p(ace)
2. p(red)
3. p(diamond)
4. p(face)
5. p(2 or 3)
6. p(black or heart)
7. p(not a king)
8. p(not a face)

there are 6 red, 3 blue, and 1 white marbles in a jar. find:

1. p(blue)
2. p(red or white)
3. p(green)
4. p(not white)
5. p(not yellow)

Explanation:

Step1: Recall probability formula

$P(E)=\frac{n(E)}{n(S)}$, where $n(E)$ is the number of favorable outcomes and $n(S)$ is the number of total outcomes.

Step2: Solve for rolling a die - P(roll a 5)

A die has 6 faces. The number of ways to roll a 5 is 1. So $P(\text{roll a }5)=\frac{1}{6}$.

Step3: Solve for rolling a die - P(roll a 1 or 6)

The number of ways to roll a 1 is 1 and the number of ways to roll a 6 is 1. So $n(E) = 2$ and $n(S)=6$. Then $P(\text{roll a }1\text{ or }6)=\frac{2}{6}=\frac{1}{3}$.

Step4: Solve for rolling a die - P(odd #)

The odd - numbered outcomes on a die are 1, 3, 5. So $n(E)=3$ and $n(S) = 6$. Then $P(\text{odd }\#)=\frac{3}{6}=\frac{1}{2}$.

Step5: Solve for rolling a die - P(multiple of 3)

The multiples of 3 on a die are 3 and 6. So $n(E)=2$ and $n(S)=6$. Then $P(\text{multiple of }3)=\frac{2}{6}=\frac{1}{3}$.

Step6: Solve for rolling a die - P(not a 4)

The number of ways to roll a 4 is 1. So the number of ways to not roll a 4 is $6 - 1=5$. Then $P(\text{not a }4)=\frac{5}{6}$.

Step7: Solve for drawing a card from a deck - P(Ace)

A standard deck has 52 cards and 4 aces. So $P(\text{Ace})=\frac{4}{52}=\frac{1}{13}$.

Step8: Solve for drawing a card from a deck - P(Red)

There are 26 red cards in a 52 - card deck. So $P(\text{Red})=\frac{26}{52}=\frac{1}{2}$.

Step9: Solve for drawing a card from a deck - P(Diamond)

There are 13 diamonds in a 52 - card deck. So $P(\text{Diamond})=\frac{13}{52}=\frac{1}{4}$.

Step10: Solve for drawing a card from a deck - P(face)

There are 12 face - cards (4 Jacks, 4 Queens, 4 Kings) in a 52 - card deck. So $P(\text{face})=\frac{12}{52}=\frac{3}{13}$.

Step11: Solve for drawing a card from a deck - P(2 or 3)

There are 4 twos and 4 threes in a 52 - card deck. So $n(E)=4 + 4=8$ and $P(2\text{ or }3)=\frac{8}{52}=\frac{2}{13}$.

Step12: Solve for drawing a card from a deck - P(Black or Heart)

There are 26 black cards and 13 hearts in a 52 - card deck. But we have double - counted the 0 overlapping cards in this case. So $P(\text{Black or Heart})=\frac{26+13}{52}=\frac{39}{52}=\frac{3}{4}$.

Step13: Solve for drawing a card from a deck - P(not a King)

There are 4 kings in a 52 - card deck. So the number of non - king cards is $52 - 4 = 48$. Then $P(\text{not a King})=\frac{48}{52}=\frac{12}{13}$.

Step14: Solve for drawing a card from a deck - P(not a face)

There are 12 face - cards in a 52 - card deck. So the number of non - face cards is $52-12 = 40$. Then $P(\text{not a face})=\frac{40}{52}=\frac{10}{13}$.

Step15: Solve for marbles - P(blue)

There are 6 red, 3 blue and 1 white marbles, so $n(S)=6 + 3+1=10$. The number of blue marbles $n(E)=3$. Then $P(\text{blue})=\frac{3}{10}$.

Step16: Solve for marbles - P(red or white)

The number of red marbles is 6 and the number of white marbles is 1. So $n(E)=6 + 1=7$ and $P(\text{red or white})=\frac{7}{10}$.

Step17: Solve for marbles - P(green)

There are 0 green marbles. So $P(\text{green})=\frac{0}{10}=0$.

Step18: Solve for marbles - P(not white)

The number of white marbles is 1. So the number of non - white marbles is $10 - 1=9$. Then $P(\text{not white})=\frac{9}{10}$.

Step19: Solve for marbles - P(not yellow)

There are 0 yellow marbles. So the number of non - yellow marbles is 10. Then $P(\text{not yellow})=\frac{10}{10}=1$.

Answer:

1. $\frac{1}{6}$
2. $\frac{1}{3}$
3. $\frac{1}{2}$
4. $\frac{1}{3}$
5. $\frac{5}{6}$
6. $\frac{1}{13}$
7. $\frac{1}{2}$
8. $\frac{1}{4}$
9. $\frac{3}{13}$
10. $\frac{2}{13}$
11. $\frac{3}{4}$
12. $\frac{12}{13}$
13. $\frac{10}{13}$
14. $\frac{3}{10}$
15. $\frac{7}{10}$
16. $0$
17. $\frac{9}{10}$
18. $1$