Question

a card is drawn randomly from a standard 52 - card deck. find the probability of the given event. write your answers as reduced fractions or whole numbers.
(a) the card drawn is 10
p(10) =
(b) the card drawn is the 10 of clubs
p(10 of clubs) =
(c) the card drawn is a face card (jack, queen, or king)
p(face card) =
(d) the card drawn is not a diamond
p(not a diamond) =
question 8
the pto is selling raffle tickets to raise money for classroom supplies. a raffle ticket costs $1. there is 1 winning ticket out of the 240 tickets sold. the winner gets a prize worth $98. round your answers to the nearest cent.
what is the expected value (to you) of one raffle ticket? $
calculate the expected value (to you) if you purchase 12 raffle tickets. $
what is the expected value (to the pto) of one raffle ticket? $

Explanation:

Step1: Calculate probability for part (a)

The formula for probability is \(P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). A standard deck has \(n = 52\) cards. There are \(4\) cards with the number \(10\) (one for each suit). So \(P(10)=\frac{4}{52}=\frac{1}{13}\).

Step2: Calculate probability for part (b)

There is only \(1\) card which is the \(10\) of clubs. So \(P(10\text{ of clubs})=\frac{1}{52}\).

Step3: Calculate probability for part (c)

There are \(3\) face - cards (Jack, Queen, King) per suit and \(4\) suits. So the number of face - cards \(m=3\times4 = 12\). Then \(P(\text{face card})=\frac{12}{52}=\frac{3}{13}\).

Step4: Calculate probability for part (d)

There are \(13\) diamond cards. The probability of getting a diamond card \(P(\text{diamond})=\frac{13}{52}\). Using the formula \(P(\text{not }A)=1 - P(A)\), we have \(P(\text{not a diamond})=1-\frac{13}{52}=\frac{39}{52}=\frac{3}{4}\).

Step5: Calculate expected value for one raffle ticket

The formula for expected value \(E(X)=\sum_{i}x_{i}P(x_{i})\). If you buy a ticket, \(x_1=(98 - 1)=97\) (net gain if you win) with \(P(x_1)=\frac{1}{240}\), and \(x_2=- 1\) (net loss if you lose) with \(P(x_2)=\frac{239}{240}\). Then \(E(X)=97\times\frac{1}{240}+(-1)\times\frac{239}{240}=\frac{97 - 239}{240}=\frac{-142}{240}\approx - 0.59\).

Step6: Calculate expected value for 12 raffle tickets

Using the property \(E(aX)=aE(X)\) (where \(a = 12\) and \(E(X)\) is the expected value of one ticket). So \(E(12X)=12\times(-\frac{142}{240})=\frac{-142\times12}{240}=\frac{-1704}{240}=-7.1\).

Step7: Calculate expected value for PTO of one raffle ticket

The PTO's perspective: if you win, the PTO's net gain is \(-97\) (gives out a \(98\) - prize but took \(1\) for the ticket), and if you lose, the PTO's net gain is \(1\). So \(E(X)_{PTO}=-97\times\frac{1}{240}+1\times\frac{239}{240}=\frac{-97 + 239}{240}=\frac{142}{240}\approx0.59\).

Answer:

(a) \(\frac{1}{13}\)
(b) \(\frac{1}{52}\)
(c) \(\frac{3}{13}\)
(d) \(\frac{3}{4}\)
Expected value of one raffle ticket: \(-0.59\)
Expected value of 12 raffle tickets: \(-7.1\)
Expected value of one raffle ticket for PTO: \(0.59\)