A hand of 5 cards is dealt to each of

Kaspaueru2 2021-12-20 Answered
A hand of 5 cards is dealt to each of three players from a standard deck of 52 cards.
(a)What is the probability that one of the players receives all four Aces?
(b)What is the probability that at least one player receives no hearts?

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Expert Answer

kaluitagf
Answered 2021-12-21 Author has 5757 answers

Step 1
Given,
A hand of 5 cards is dealt to each of three players from a standard deck of 52 cards.
Step 2
(a) The probability that one of the players receives all four Aces:
Let X be the number of Ace cards that a player receives
The probability of getting an Ace is \(\displaystyle{p}={\frac{{{4}}}{{{52}}}}={0.0769}\)
\(\displaystyle{n}={5}\)
\(\displaystyle{X}\sim{B}in{o}{m}{i}{a}{l}{\left({5},{0.0769}\right)}\)
\(\displaystyle{P}{\left({X}={4}\right)}={5}{C}_{{{4}}}{\left({0.0769}\right)}^{{{4}}}{\left({1}-{0.0769}\right)}^{{{5}-{4}}}\)
\(\displaystyle={0.00016}\)
(b) The probability that at least one player receives no hearts :
Let X be the number of heart cards that a player receives
The probability of getting a heart is \(\displaystyle{p}={\frac{{{13}}}{{{52}}}}={0.25}\)
\(\displaystyle{n}={5}\)
\(\displaystyle{X}\sim{B}in{o}{m}{i}{a}{l}{\left({5},{0.25}\right)}\)
The probability that no player receives no heart :
\(\displaystyle{P}{\left({X}={0}\right)}={5}{C}_{{{0}}}{\left({0.25}\right)}^{{{0}}}{\left({1}-{0.25}\right)}^{{{5}-{0}}}\)
\(\displaystyle={0.2373}\)
The probability that at least one player receives no hearts \(\displaystyle={1}-\text{No player receives no heart}\)
\(\displaystyle={1}-{0.2373}\)
\(\displaystyle={0.7627}\)

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Lynne Trussell
Answered 2021-12-22 Author has 5647 answers

Use inclusion-exclusion. Call the players A, B, and C.
The probability of only one player getting two aces will be 3 times the probability of any one of them getting two aces minus the instances where two of the players get two aces:
\(\displaystyle{P}{\left({A}\right)}+{P}{\left({B}\right)}+{P}{\left({C}\right)}-{\left\lbrace{P}{\left({A}{B}\right)}+{P}{\left({A}{C}\right)}+{P}{\left({B}{C}\right)}\right\rbrace}\)
So,
\(3\times\frac{(4\ choose\ 2)(48\ choose\ 3)}{52\ choose\ 5}-3\times\frac{(4\ choose\ 2)(48\ choose\ 3.3,\ and 42)}{52\ choose\ 5.5,\ and\ 42}\)

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nick1337
Answered 2021-12-28 Author has 10160 answers

There 52 cards and 13 cards in each suit (clubs,spades,heart and diamonds).
We have 13 hearts and 39 other cards. We will find the probability of being dealt at least one heart by using the complement rule. Note the following:
\(P\text{(being dealt at least one heart)} = 1 - P\text{(being dealt no hearts)}\)
\(=1-\frac{((39),(5))}{((52),(5))}\)
\(\approx0.7785\)
The probability that a player is dealt 5 cards with at least one heart is is approximately a 0.7785.

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