(a) P (odd number)
(d) P (multiple of 4 )
(e) P (at least 3 )
2. If a card is drawn at random from a deck of 52 well shuffled playing cards, find the probability that the card is:
© an ace
(a) red
(b) a club
(d) a red queen
(e) a face card
3. If a pair of dice is thrown once, find each of the following.
(a) sum 7
(b) sum 12)
© sum
(d) P (even number on first dice)
4. The adjoining spinner is rotated once. Find each of the following.
(a) P (number 7 )
(b) P (even number)
© P (multiple of 3 )
(d) P (factor of 28 )

Ask by Welch French. in India

Mar 21,2025

Question

(a) P (odd number)
(d) P (multiple of 4 )
(e) P (at least 3 )
2. If a card is drawn at random from a deck of 52 well shuffled playing cards, find the probability that the card is:
© an ace
(a) red
(b) a club
(d) a red queen
(e) a face card
3. If a pair of dice is thrown once, find each of the following.
(a) sum 7
(b) sum 12)
© sum
(d) P (even number on first dice)
4. The adjoining spinner is rotated once. Find each of the following.
(a) P (number 7 )
(b) P (even number)
© P (multiple of 3 )
(d) P (factor of 28 )

Ask by Welch French. in India

Mar 21,2025

UpStudy AI · Accepted Answer

Let's solve each part of the problem step by step.

### 1. Probability of Events Related to a Single Die Roll

(a) **P(odd number)**: The odd numbers on a die are 1, 3, and 5. There are 3 odd numbers out of 6 total numbers.

$$
P(\text{odd number}) = \frac{3}{6} = \frac{1}{2}
$$

(d) **P(multiple of 4)**: The multiples of 4 on a die are only 4. There is 1 multiple of 4 out of 6 total numbers.

$$
P(\text{multiple of 4}) = \frac{1}{6}
$$

(e) **P(at least 3)**: The numbers that are at least 3 are 3, 4, 5, and 6. There are 4 such numbers.

$$
P(\text{at least 3}) = \frac{4}{6} = \frac{2}{3}
$$

### 2. Probability of Events Related to a Deck of Cards

(c) **P(an ace)**: There are 4 aces in a deck of 52 cards.

$$
P(\text{an ace}) = \frac{4}{52} = \frac{1}{13}
$$

(a) **P(red)**: There are 26 red cards (hearts and diamonds) in a deck of 52 cards.

$$
P(\text{red}) = \frac{26}{52} = \frac{1}{2}
$$

(b) **P(a club)**: There are 13 clubs in a deck of 52 cards.

$$
P(\text{a club}) = \frac{13}{52} = \frac{1}{4}
$$

(d) **P(a red queen)**: There are 2 red queens (one from hearts and one from diamonds).

$$
P(\text{a red queen}) = \frac{2}{52} = \frac{1}{26}
$$

(e) **P(a face card)**: There are 12 face cards (3 face cards in each of the 4 suits).

$$
P(\text{a face card}) = \frac{12}{52} = \frac{3}{13}
$$

### 3. Probability of Events Related to a Pair of Dice

(a) **P(sum 7)**: The combinations that give a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 combinations.

$$
P(\text{sum 7}) = \frac{6}{36} = \frac{1}{6}
$$

(b) **P(sum 12)**: The only combination that gives a sum of 12 is (6,6). There is 1 combination.

$$
P(\text{sum 12}) = \frac{1}{36}
$$

(c) **P(sum $$\geq 8$$)**: The combinations that give a sum of 8 or more are: 
- Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 combinations
- Sum 9: (3,6), (4,5), (5,4), (6,3) → 4 combinations
- Sum 10: (4,6), (5,5), (6,4) → 3 combinations
- Sum 11: (5,6), (6,5) → 2 combinations
- Sum 12: (6,6) → 1 combination

Total combinations for sum $$\geq 8$$ = 5 + 4 + 3 + 2 + 1 = 15.

$$
P(\text{sum} \geq 8) = \frac{15}{36} = \frac{5}{12}
$$

(d) **P(even number on first die)**: The even numbers on a die are 2, 4, and 6. There are 3 even numbers out of 6.

$$
P(\text{even number on first die}) = \frac{3}{6} = \frac{1}{2}
$$

### 4. Probability of Events Related to a Spinner

Assuming the spinner has numbers from 1 to 10 (for example):

(a) **P(number 7)**: There is 1 occurrence of the number 7.

$$
P(\text{number 7}) = \frac{1}{10}
$$

(b) **P(even number)**: The even numbers are 2, 4, 6, 8, and 10. There are 5 even numbers.

$$
P(\text{even number}) = \frac{5}{10} = \frac{1}{2}
$$

(c) **P(multiple of 3)**: The multiples of 3 are 3, 6, and 9. There are 3 multiples of 3.

$$
P(\text{multiple of 3}) = \frac{3}{10}
$$

(d) **P(factor of 28)**: The factors of 28 are 1, 2, 4, 7, 14, and 28. Assuming the spinner has numbers from 1 to 10, the factors of 28 that are within this range are 1, 2, and 4. There are 3 factors.

$$
P(\text{factor of 28}) = \frac{3}{10}
$$

### Summary of Probabilities
1. 
   - P(odd number) = $$ \frac{1}{2} $$
   - P(multiple of 4) = $$ \frac{1}{6} $$
   - P(at least 3) = $$ \frac{2}{3} $$

2. 
   - P(an ace) = $$ \frac{1}{13} $$
   - P(red) = $$ \frac{1}{2} $$
   - P(a club) = $$ \frac{1}{4} $$
   - P(a red queen) = $$ \frac{1}{26} $$
   - P(a face card) = $$ \frac{3}{13} $$

3. 
   - P(sum 7) = $$ \frac{1}{6} $$
   - P(sum 12) = $$ \frac{1}{36} $$
   - P(sum $$\geq 8$$) = $$ \frac{5}{12} $$
   - P(even number on first die) = $$ \frac{1}{2} $$

4. 
   - P(number 7) = $$ \frac{1}{10} $$
   - P(even number) = $$ \frac{1}{2} $$
   - P(multiple of 3) = $$ \frac
1. - P(odd number) = 1/2 - P(multiple of 4) = 1/6 - P(at least 3) = 2/3 2. - P(an ace) = 1/13 - P(red) = 1/2 - P(a club) = 1/4 - P(a red queen) = 1/26 - P(a face card) = 3/13 3. - P(sum 7) = 1/6 - P(sum 12) = 1/36 - P(sum ≥8) = 5/12 - P(even number on first die) = 1/2 4. - P(number 7) = 1/10 - P(even number) = 1/2 - P(multiple of 3) = 3/10 - P(factor of 28) = 3/10

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