/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A deck of 52 cards is mixed well... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 25

A deck of 52 cards is mixed well, and 5 cards are dealt. a. It can be shown that (disregarding the order in which the cards are dealt) there are \(2,598,960\) possible five-card hands, of which only 1287 are hands consisting entirely of spades. What is the probability that a hand will consist entirely of spades? What is the probability that a hand will consist entirely of a single suit? b. It can be shown that exactly 63,206 hands contain only spades and clubs, with both suits represented. What is the probability that a hand consists entirely of spades and clubs with both suits represented? c. Using the result of Part (b), what is the probability that a hand contains cards from exactly two suits?

Short Answer

Expert verified
The probability of drawing a hand that is only spades is approximately \(0.000495\) and the probability of drawing a hand from any single suit is about \(0.00198\). The probability that a hand consists of both spades and clubs is approximately \(0.02431\). The probability that a hand contains exactly two suits is around \(0.14586\)

Step by step solution

01

Probability of a hand with only spades and single suit

For the first part, we know that there are \(2,598,960\) total possible hands and that 1287 of them consist entirely of spades. We can calculate the probability of pulling a hand of all spades by dividing the number of hands with all spades by the total number of possible hands: \(\frac{1287}{2,598,960}\). To find the probability that a hand will consist entirely of a single suit, we need to multiply the probability of getting a hand of spades by 4 (since there are 4 possible suits: hearts, diamonds, clubs, and spades)
02

Probability of a hand with both spades and clubs

For the second part, we are told that 63,206 hands contain only spades and clubs, with both suits represented. We can calculate the probability by dividing the favorable outcomes by the total number of possible outcomes: \(\frac{63,206}{2,598,960}\)
03

Probability of a hand with exactly two suits

For the third part, we can use the result from part b to find the probability that a hand contains cards from exactly two suits. Since there are six possible pairs of two different suits (spades-hearts, spades-diamonds, spades-clubs, hearts-diamonds, hearts-clubs, and diamonds-clubs), we multiply the probability found in part b by 6

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Most popular questions from this chapter

The newspaper article "Folic Acid Might Reduce Risk of Down Syndrome" (USA Today, September 29 . 1999) makes the following statement: "Older women are at a greater risk of giving birth to a baby with Down Syndrome than are younger women. But younger women are more fertile, so most children with Down Syndrome are born to mothers under \(30 . "\) Let \(D=\) event that a randomly selected baby is born with Down Syndrome and \(Y=\) event that a randomly selected baby is born to a young mother (under age 30 ). For each of the following probability statements, indicate whether the statement is consistent with the quote from the article, and if not, explain why not. a. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.004, P(Y)=.7\) b. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.001, P(Y)=.7\) c. \(P(D \mid Y)=.004, P\left(D \mid Y^{C}\right)=.004, P(Y)=.7\) d. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.004, P(Y)=.4\) e. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.001, P(Y)=.4\) f. \(P(D \mid Y)=.004, P\left(D \mid Y^{C}\right)=.004, P(Y)=.4\)

In a school machine shop, \(60 \%\) of all machine breakdowns occur on lathes and \(15 \%\) occur on drill presses. Let \(E\) denote the event that the next machine breakdown is on a lathe, and let \(F\) denote the event that a drill press is the next machine to break down. With \(P(E)=.60\) and \(P(F)=.15\), calculate: a. \(P\left(E^{C}\right)\) b. \(P(E \cup F)\) c. \(P\left(E^{C} \cap F^{C}\right)\)

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An article in the New york Times (March 2. 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.
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