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Question: What is the probability that a five-card hand contains two pairs (that is, two of each of two different kinds and a fifth card of a third kind)?

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Answer

The probability that a five-card hand contains two pairs (that is, each two of different kind and a fifth card of a third kind) isP(E)=0.0475.

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01

 Given  

A pack of cards. In a pack of cards there are 52cards

02

The Concept of Probability

IfSrepresents the sample space andErepresents the event. Then the probability of occurrence of favourable event is given by the formula as below:

P(E)=n(E)n(S)

03

Determine the probability

Here, we have to find the probability that a five-card poker hand contain contains two pairs (that is, each two of different kind and a fifth card of a third kind).

If Srepresents the sample space and Erepresents the event. Then the probability of occurrence of favourable event is given by the formula as below:

P(E)=n(E)n(S).

As we know that there a total of fifty-two cards and five cards can be chosen in52C5.

n(S)=52C5

We are supposed to draw a five-card poker hand contains two pairs (that is, each two of different kind and a fifth card of a third kind) this can be accomplished as follows:

Number of ways to choose two pairs (that is, each two of different kind and a fifth card of a third kind)=C(13,2)×C(4,2)×C(4,2)×C(44,1)

P(E)=C(13,2)×C(4,2)×C(4,2)×C(44,1)C552P(E)=13!2!×11!×4!2!×2!×4!2!×2!×44!1!×43!52!47!×5!P(E)=1984165P(E)=0.0475

The probability that a five-card hand contains two pairs (that is, each two of different kind and a fifth card of a third kind) isP(E)=0.0475 .

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Most popular questions from this chapter

Question: Find the probability of winning a lottery by selecting the correct six integers, where the order in which these integers are selected does not matter, from the positive integers not exceeding.

(a) 30.

(b) 36.

(c) 42.

(d) 48.

Question: Prove Theorem \(2\), the extended form of Bayes’ theorem. That is, suppose that \(E\) is an event from a sample space \(S\) and that \({F_1},{F_2},...,{F_n}\) are mutually exclusive events such that \(\bigcup\nolimits_{i = 1}^n {{F_i} = S} \). Assume that \(p\left( E \right) \ne 0\) and \(p\left( {{F_i}} \right) \ne 0\) for \(i = 1,2,...,n\). Show that

\(p\left( {{F_j}\left| E \right.} \right) = \frac{{p\left( {E\left| {{F_j}} \right.} \right)p\left( {{F_j}} \right)}}{{\sum\nolimits_{i = 1}^n {p\left( {E\left| {{F_i}} \right.} \right)p\left( {{F_i}} \right)} }}\)

(Hint: use the fact that \(E = \bigcup\nolimits_{i = 1}^n {\left( {E \cap {F_i}} \right)} \).)

Question: An electronics company is planning to introduce a new camera phone. The company commissions a marketing report for each new product that predicts either the success or the failure of the product. Of new products introduced by the company, 60% have been successes. Furthermore. 70% of their successful products were predicted to be successes, while 40% of failed products were predicted to be successes. Find the probability that this new camera phone will be successful if its success has been predicted.

Suppose that a test for opium use has a 2% false positive rate and a 5% false negative rate. That is, 2% of people who do not use opium test positive for opium, and 5% of opium users test negative for opium. Furthermore, suppose that 1% of people actually use opium.

a)Find the probability that someone who tests negative for opium use does not use opium.

b) Find the probability that someone who tests positive for opium use actually uses opium.

Question: What is the probability that a five-card poker hand contain the two of diamonds and three of spades?

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