Question

Probability and OddsAnswer all questions with a fraction in lowest terms. If you’d like, you can also write them as a percent. If you draw one card at random, what is the probability that card is a(n)…9.Heart?10.7 of diamonds?11.face card or a club?Given the card is a club, what is the probability a card drawn at random will be a(n)…12.8?13.10 or ace?You are choosing two cards, without replacing the first card. What is the probability you choose…14.a 7 then a 3?15.two consecutive fours?16.two consecutive diamonds?You are choosing two cards, replacing the first card in the deck after it has been drawn. What is the probability you choose…17.a 7 then a 3?18.two consecutive fours?19.two consecutive diamonds?

Answer 1

Step-by-step explanation

We must answer the following question: If you draw one card at random, what is the probability that card is:

• 9. Heart?

,

• 10. 7 of diamonds?

,

• 11. face card or a club?

The total number of cards in the deck is:

• n(total) = 52.

9) The number of hearts in the deck is:

• n(heart) = 13.

The probability of drawing a heart at random is:

`P(heart)=(n(heart))/(n(total))=(13)/(52)=(1)/(4).`

10) The number of 7 of diamonds in the deck is:

• n(7 of diamonds) = 1.

The probability of drawing a 7 of diamonds is:

`P(7\text{ of diamonds})=\frac{n(7\text{ of diamonds})}{n(total)}=(1)/(52).`

11) We have the following number of cards in the deck:

• n(face card) = 3 x 4 = 12,

,

• n(club) = 13,

,

• n(face card and club) = 3.

The probability of drawing a face card or a club is given by:

`\begin{gathered} P(\text{face card or club})=\frac{n(\text{face card})+n(\text{club})-n(\text{face card and club})}{n(total)} \\ =(12+13-3)/(52)=(22)/(52)=(11)/(26). \end{gathered}`Answer

9) P(heart) = 1/4

10) P(7 of diamonds) = 1/52

11) P (face card or a club) = 11/26