Compute the probability of randomly drawing five cards from a deck and getting exactly two aces.
An alternative way to do this is to use equations for combinations, the general formula for which is: Show #C_(n,k)=(n!)/((k)!(n-k)!)# with #n="population", k="picks"# We want our hand to have exactly one Ace. Since there are four aces in a deck, we can set #n=4, k=1#, and so: #C_(4,1)# But we also need to account for the other four cards in our hand. There are 48 cards to pick from, so we can set #n=48, k=4#, and so: #C_(48,4)# We multiply them together to find the total number of ways we can get exactly one Ace in our hand: #C_(4,1)xxC_(48,4)=(4!)/((1)!(4-1)!)(48!)/((4)!(48-4)!)=># #(cancel(4!)(48!))/((3!)cancel(4!)(44!))=(cancel(48)^8xx47xx46xx45xxcancel44!)/(cancel(3xx2)xxcancel44!)=># #8xx47xx46xx45=778,320# ~~~~~ To figure out the probability, we also need to know the total number of 5-card hands possible: #C_(52,5)=(52!)/((5)!(52-5)!)=(52!)/((5!)(47!))# Let's evaluate it! #(52xx51xxcancelcolor(orange)(50)^10xx49xxcancelcolor(red)48^2xxcancelcolor(brown)(47!))/(cancelcolor(orange)5xxcancelcolor(red)(4xx3xx2)xxcancelcolor(brown)(47!))=52xx51xx10xx49xx2=2,598,960# And so: #P=(778,320)/(2,598,960)~=29.9%# We can use permutations and combinations to help us answer more complex probability questions Example 1 A 4 digit PIN is selected. What is the probability that there are no repeated digits? 10P4 104=504010000=0.504\frac{{{}_{{10}}{P}_{{4}}}}{{{10}^{{4}}}}=\frac{{5040}}{{10000}}={0.504} Example 2 In a certain state's lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a
player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn in doesn't matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket. 6C648C 6=112271512≈=0.0000000815\frac{{{}_{{6}}{C}_{{6}}}}{{{}_{{48}}{C}_{{6}}}}=\frac{{1}}{{12271512}}\approx={0.0000000815} Example 3 In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a
second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket. (6C5)(42C1)48C6=25212271512≈0.0000205\frac{{{\left({}_{{6}}{C}_{{5}}\right)}{\left({}_{{42}}{C}_{{1}}\right)}}}{{{}_{{48}}{C}_{{6}}}}=\frac{{252}}{{12271512}}\approx{0.0000205} Try it Now A multiple-choice question on an economics quiz contains 10 questions with five possible answers each. Compute the probability of randomly guessing the answers and getting 9 questions correct. Example 4 Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace. P(one Ace)=(4C1)(48C4) 52C5=7783202598960≈0.299{P}{\left(\text{one Ace}\right)}=\frac{{{\left({}_{{4}}{C}_{{1}}\right)}{\left({}_{{48}}{C}_{{4}}\right)}}}{{{}_{{52}}{C}_{{5}}}}=\frac{{778320}}{{2598960}}\approx{0.299} Example 5 Compute the probability of
randomly drawing five cards from a deck and getting exactly two Aces. P(two Ace)=(4 C2)(48C3)52C5=1037762598960≈0.0399{P}{\left(\text{two Ace}\right)}=\frac{{{\left({}_{{4}}{C}_{{2}}\right)}{\left({}_{{48}}{C}_{{3}}\right)}}}{{{}_{{52}}{C}_{{5}}}}=\frac{{103776}}{{2598960}}\approx{0.0399} It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier. Try it Now Compute the probability of randomly drawing five cards from a deck of cards and getting three Aces and two Kings. Birthday ProblemLet's take a pause to consider a famous problem in probability theory: Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday? Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30/365, perhaps?). Let's see if we should listen to our intuition. Let's start with a simpler problem, however. Example 6 Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people? P(no shared birthday)=365365⋅364365⋅363365≈0.9918{P}{\left(\text{no shared birthday}\right)}=\frac{{365}}{{365}}\cdot\frac{{364}}{{365}}\cdot\frac{{363}}{{365}}\approx{0.9918} and then subtract from 1 to get This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small. Let's make our group a bit bigger. Example 7 Suppose five people are in a room. What is the probability that there is at least one shared birthday among these five people? P(shared birthday)=1−365365⋅364365⋅363365⋅362365⋅361365 ≈0.0271{P}{\left(\text{shared birthday}\right)}={1}-\frac{{365}}{{365}}\cdot\frac{{364}}{{365}}\cdot\frac{{363}}{{365}}\cdot\frac{{362}}{{365}}\cdot\frac{{361}}{{365}}\approx{0.0271} Note that we could rewrite this more compactly as P(shared birthday )=1−365P53655≈0.0271{P}{\left(\text{shared birthday}\right)}={1}-\frac{{{}_{{365}}{P}_{{5}}}}{{365}^{{5}}}\approx{0.0271} which makes it a bit easier to type into a calculator or computer,
and which suggests a nice formula as we continue to expand the population of our group. Example 8 Suppose 30 people are in a room. What is the probability that there is at least one shared birthday among these 30 people? P(shared birthday)=1− 365P3036530≈0.706{P}{\left(\text{shared birthday}\right)}={1}-\frac{{{}_{{365}}{P}_{{30}}}}{{365}^{{30}}}\approx{0.706} which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least
one shared birthday! If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn't studied probability!) You wouldn't be guaranteed to win, but you should win more than half the time. Try it Now Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people? Expected ValueExpected value is perhaps the most useful probability concept we will discuss. It has many applications, from insurance policies to making financial decisions, and it's one thing that the casinos and government agencies that run gambling operations and lotteries hope most people never learn about.Example 9 In the casino game roulette, a wheel with 38 spaces (18 red, 18 black, and 2 green) is spun. In one possible bet, the player bets $1 on a single number. If that number is spun on the
wheel, then they receive $36 (their original $1 + $35). Otherwise, they lose their $1. On average, how much money should a player expect to win or lose if they play this game repeatedly? We call this average gain or loss the expected value of playing roulette. Notice that no one ever loses exactly 5.3 cents: most people (in fact, about 37 out of every 38) lose $1 and a very few people (about 1 person out of every 38) gain $35 (the $36 they win minus the $1 they spent to play the game). 138\frac{1}{38} . The complement, the probability of losing, is 3738\frac{37}{38} .
Notice that if we multiply each outcome by its corresponding probability we get $35 × 138\frac{{1}}{{38}} = 0.9211 and –$1 × 3738 \frac{37}{38} = –0.9737, and if we add these numbers we get 0.9211 + (–0.9737) ≈ –0.053, which is the expected value we computed above. Expected ValueExpected Value is the average gain or loss of an event if the procedure is repeated many times. Try it Now You purchase a raffle ticket to help out a charity. The raffle ticket costs $5. The charity is selling 2000 tickets. One of them will be drawn and the person holding the ticket will be given a prize worth $4000. Compute the expected value for this raffle. Example 10 In a certain state's lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers
drawn match the numbers that a player had chosen, the player wins $1,000,000. If they match 5 numbers, then win $1,000. It costs $1 to buy a ticket. Find the expected value. 6C648 C6=112271512≈0.0000000815\frac{{{}_{{6}}{C}_{{6}}}}{{{}_{{48}}{C}_{{6}}}}=\frac{{1}}{{12271512}}\approx{0.0000000815} for all 6 numbers, ( 6C5)(42C1)48C6=25212271512≈0.0000205\frac{{{\left({}_{{6}}{C}_{{5}}\right)}{\left({}_{{42}}{C}_{{1}}\right)}}}{{{}_{{48}}{C}_{{6}}}}=\frac{{252}}{{12271512}}\approx{0.0000205} for 5 numbers.
The expected value, then is: ($999,999)×112271512+($999)×25212271512 +(−$1)×1227125912271512≈−$0.898{\left(\${999},{999}\right)}\times\frac{{1}}{{12271512}}+{\left(\${999}\right)}\times\frac{{252}}{{12271512}}+{\left(-\${1}\right)}\times\frac{{12271259}}{{12271512}}\approx-\${0.898} On average, one can expect to lose about 90 cents on a lottery ticket. Of course, most players will lose $1. In general, if the expected value of a game is negative, it is not a good idea to play the game, since on average you will lose money. It would be better to play a game with a positive expected value (good luck trying to find one!), although keep in mind that even if the average winnings are positive it could be
the case that most people lose money and one very fortunate individual wins a great deal of money. If the expected value of a game is 0, we call it a fair game, since neither side has an advantage. Try it Now A friend offers to play a game, in which you roll 3 standard 6-sided dice. If all the dice roll different values, you give him $1. If any two dice match values, you get $2. What is the expected value of this game? Would you play? Expected value also has applications outside of gambling. Expected value is very common in making insurance decisions. Example 11 According to the estimator at numericalexample.com, a 40-year-old man in the US has a 0.242% risk of dying during the next year. An insurance company charges $275 for a life-insurance policy that pays a $100,000 death benefit. What is the expected value for the person buying the insurance?
The expected value is ($99,725)(0.00242) + (–$275)(0.99758) = –$33. Not surprisingly, the expected value is negative; the insurance company can only afford to offer policies if they, on average, make money on each policy. They can afford to pay out the occasional benefit because they offer enough policies that those benefit payouts are balanced by the rest of the insured people. What is the probability of drawing 2 aces from a deck of cards?The chance of drawing one of the four aces from a standard deck of 52 cards is 4/52; but the chance of drawing a second ace is only 3/51, because after we drew the first ace, there were only three aces among the remaining 51 cards. Thus, the chance of drawing an ace on each of two draws is 4/52 × 3/51, or 1/221.
What is the probability of randomly draw five cards from a deck of cards and getting exactly one ace?Explanation: There are 4 aces in the 52-card deck so the probability of dealing an ace is 4/52 = 1/13. In a 5-card hand, each card is equally likely to be an ace with probability 1/13. So together, the expected number of aces in a 5-card hand is 5 * 1/13 = 5/13.
What is the probability that a 5 is drawn from a deck of cards?So drawing a five-card hand of a single selected suit is a rare event with a probability of about one in 2000.
How many 52 Answers. There are 10 possible 5 -card hands with exactly 3 kings and exactly 2 aces.
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