Question

Understanding the Concepts and Skills 5.77 Playing Cards. An ordinary deck of playing cards has 52 cards. There are four suits-spades, hearts, diamonds, and clubs with 13 cards in each suit. Spades and clubs are black; hearts and dia- monds are red. One of these cards is selected at random. Let R denote the event that a red card is chosen. Find the probability that a red card is chosen, and express your answer in probability notation. 5.78 Poker Chips. A bowl contains 12 poker chips3 red, 4 white, and 5 blue. One of these poker chips is selected at random from the bowl. Let B denote the event that the chip selected is blue. Find the probability that a blue chip is selected, and express your answer in probability notation 5.79 Suppose that A and B are mutually exclusive events such that P(A) = 0.25 and P(B) = 0.40. Determine PCA or B). 5.80 Suppose that and Dare mutually exclusive events such that P(C) = 0.14 and P(D) = 0.32. Determine PC or D). 5.81 Let E be an event with probability 0.35. Find the probability of (not E). 5.82 Let F be an event with probability 0.72. Find the probability of (not F). 5.83 Suppose that C and D are events such that P(C)=0.35, P(D)=0.40, and PC & D) = 0.30. Determine P(C or D). 5.84 Suppose that A and B are events such that P(A) = 0.84. P(B) =0.46, and P(A&B) = 0.38. Determine PCA or B). 5.85 Suppose that A and B are events such that P(A) = 1/3. P(A or B) = 1/2, and P(A&B) = 1/10. a. Are events A and B mutually exclusive? Explain your answer. b. Find P(B) 5.86 Suppose that A and B are events such that P(A)= 1/4, P(B) = 1/3, and P(A or B) = 1/2 a. Are events A and B mutually exclusive? Explain your answer. b. Find P(A&B). S SC Applying the Concepts and Skills 5.87 Ages of Senators. According to the Congressional Diree tory, the official directory of the U.S. Congress, prepared by the Joint Committee on Printing, the age distribution for senators in the U.S. Congress as of Fall 2013, is as shown in the following table. d. Age lyn No. of senators Under 50 50-59 60-69 11 30 5.89 Abu

Answer 1

`\( P(R) = (26)/(52) = (1)/(2) \)`

The probability of choosing a blue chip is
`\( (5)/(12) \).`

The probability of either event A or event B occurring is
`\( 0.65 \).`

Step-by-step explanation:

In a standard deck of 52 playing cards, there are 26 red cards (13 hearts and 13 diamonds) out of a total of 52 cards. Therefore, the probability of selecting a red card, denoted by P(R, is given by the ratio of the number of red cards to the total number of cards:

`\[ P(R) = \frac{\text{Number of Red Cards}}{\text{Total Number of Cards}} = (26)/(52) = (1)/(2) \]`

So, the probability of choosing a red card is
`\( (1)/(2) \)`.

In the bowl with 12 poker chips, there are 5 blue chips. The probability of selecting a blue chip, denoted by P(B) , is given by the ratio of the number of blue chips to the total number of chips in the bowl:

`\[ P(B) = \frac{\text{Number of Blue Chips}}{\text{Total Number of Chips}} = (5)/(12) \]`

Therefore, the probability of choosing a blue chip is
`\( (5)/(12) \).`

When events A and B are mutually exclusive, it means that they cannot occur simultaneously. In such cases, the probability of the union of these events
`(\( P(A \cup B) \))` is simply the sum of their individual probabilities (\( P(A) + P(B) \)).

Given that P(A) = 0.25 and ( P(B) = 0.40 ), the probability of the union \(
`P(A \cup B) \)` is calculated as:

`\[ P(A \cup B) = P(A) + P(B) = 0.25 + 0.40 = 0.65 \]`

Therefore, the probability of either event A or event B occurring is
`\( 0.65 \).`

5.81
`\( P(C \cup D) = P(C) + P(D) = 0.14 + 0.32 = 0.46 \)`

For mutually exclusive events C and D, the probability of their union
`\( P(C \cup D) \)` is calculated as the sum of their individual probabilities. Given \( P(C) = 0.14 \) and \( P(D) = 0.32 \):

`\[ P(C \cup D) = P(C) + P(D) = 0.14 + 0.32 = 0.46 \]`

Therefore, the probability of either event C or event D occurring is (0.46).

5.82
`\( P(\\eg E) = 1 - P(E) = 1 - 0.35 = 0.65 \)`

The probability of the complement of event
`E (\( \\eg E \))` is calculated as one minus the probability of event E. Given
`\( P(E) = 0.35 \):`

`\[ P(\\eg E) = 1 - P(E) = 1 - 0.35 = 0.65 \]`

So, the probability of not event E occurring is ( 0.65 ).

5.83
`\( P(C \cap D) = 0 \.30 \)`

For events C and D, the probability of their intersection
`(\( P(C \cap D) \))` is given as \( 0.30 \). This represents the probability that both events C and D occur simultaneously.

`\( P(C \cup D) = P(C) + P(D) - P(C \cap D) = 0.14 + 0.32 - 0.30 = 0.16 \)`

To find the probability of the union of events C and D
`(\( P(C \cup D) \)),`we use the inclusion-exclusion principle:

`\[ P(C \cup D) = P(C) + P(D) - P(C \cap D) \]`

Given \( P(C) = 0.14 \), \( P(D) = 0.32 \), and
`\( P(C \cap D) = 0.30 \):`

`\[ P(C \cup D) = 0.14 + 0.32 - 0.30 = 0.16 \]`

So, the probability of either event C or event D occurring is \( 0.16 \).

a. Events A and B are not mutually exclusive because
`\( P(A \cap B)` = 1/10 \) (not equal to zero).

b. To find \( P(B) \), we use the formula
`\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \),` where
`\( P(A \cup B) = 1/2 \), \( P(A) = 1/3 \),` and
`\( P(A \cap B) = 1/10 \).` Solving for \( P(B) \):

`\[ P(B) = P(A \cup B) - P(A) + P(A \cap B) = (1)/(2) - (1)/(3) + (1)/(10) = (3)/(10) \]`

So,
`\( P(B) = (3)/(10) \).`

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`\[ P(A \cap B) = P(A) + P(B) - P(A \cup B) = (1)/(4) + (1)/(3) - (1)/(2) = (1)/(12) \]`

So,
`\( P(A \cap B) = (1)/(12) \).`