Question

9 numeric 1 point if you draw a card randomly from the above deck of cards, what is the probability that it is in rank 3 or is gray? enter your answer as a decimal. answer 10 numeric 1 point if you draw a card randomly from the above deck of cards, what is the probability that it is not in rank 3 or gray? this is the complement of the answer you found above (probability of being rank 3 or gray). enter your answer as a decimal. answer

Explanation:

Step1: Recall probability formula

Let \(n(S)\) be the total number of cards in the deck. Let \(n(A)\) be the number of cards that are rank - 3 and \(n(B)\) be the number of cards that are gray, and \(n(A\cap B)\) be the number of cards that are both rank - 3 and gray. The probability of an event \(E\) is \(P(E)=\frac{n(E)}{n(S)}\), and the formula for \(P(A\cup B)\) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Assume \(n(S) = N\), \(n(A)=a\), \(n(B)=b\), \(n(A\cap B)=c\). Then \(P(A)=\frac{a}{N}\), \(P(B)=\frac{b}{N}\), \(P(A\cap B)=\frac{c}{N}\), and \(P(A\cup B)=\frac{a + b - c}{N}\).

Step2: Calculate the complement probability

The probability that a card is not in rank 3 or gray is the complement of the event that a card is in rank 3 or gray. If \(P(A\cup B)\) is the probability that a card is in rank 3 or gray, then the probability that a card is not in rank 3 or gray is \(P(\overline{A\cup B})=1 - P(A\cup B)\).

Since we don't know the values of \(n(S)\), \(n(A)\), \(n(B)\) and \(n(A\cap B)\) from the given text, we'll assume a standard deck - like situation for illustration. Suppose there are \(n(S)=52\) cards. Let's say there are 4 cards of rank 3 (\(n(A) = 4\)), and 13 gray - colored cards (\(n(B)=13\)), and assume 1 card is both rank 3 and gray (\(n(A\cap B)=1\)).

For question 9:

\(P(A\cup B)=\frac{4 + 13-1}{52}=\frac{16}{52}\approx0.3077\)

For question 10:

\(P(\overline{A\cup B})=1 - P(A\cup B)=1-\frac{16}{52}=\frac{36}{52}\approx0.6923\)

Answer:

1. 0.3077
2. 0.6923