What is the probability that the deal of a five-card

Oberlaudacu 2022-01-17 Answered
What is the probability that the deal of a five-card hand provides no aces?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Robert Pina
Answered 2022-01-17 Author has 42 answers
(48 choose 5) / (52 choose 5) = 0.6588
Explanation:
All the combinations of hands that could contain no aces is the equivalent of removing the four aces from the deck and dealing all the combinations of hands from the remaining 48 cards. So, the numerator of the probability ratio should be (48 choose 5).
If you want the probability that any possible 5 card hand is a hand that contains no aces, then the denominator should contain all possible hands that can be dealt from a full deck of cards, which is (52 choose 5).
Then divide all the ''no aces'' combinations by all possible combinations of the full deck to determine the chance that any given hand dealt won't contain any aces.
Not exactly what you’re looking for?
Ask My Question
Piosellisf
Answered 2022-01-18 Author has 40 answers
Explanation:
Method 1
If we start with a full deck there are 52 cards; 4 aces; 48 non-aces
Card 1 deal should not be an Ace (48 non aces; 52 cards)
P(Card 1 not Ace)=4852
Card 2 deal should not be an Ace (47 non aces; 51 cards)
P(Card 2 not Ace)=4751
Card 3 deal should not be an Ace (46 non aces; 50 cards)
P(Card 3 not Ace)=4650
Card 4 deal should not be an Ace (45 non aces; 49 cards)
P(Card 4 not Ace)=4549
Card 5 deal should not be an Ace (44 non aces; 48 cards)
P(Card 5 not Ace)=4448
So then:
P(all five cards non aces)=48524751465045494448
=3567354145
=0.658841...
Method 2
Using the combination formula:
{n}Cr=(beg{array}{c}nrend{array})=n!r!(nr)!
The number of combination of choosing five non-aces from 52 cards (48 of which are not aces) is given by:
n(all five cards non aces)=48C5=(beg{array}{c}485end{array})
=48!5!(485)!
=48!5!43!
And the total number of all combinations of choosing any five cards
n(any five cards) =52C5=(beg{array}{c}525end{array})
=52!5!(525)!
=52!5!47!
P (all five cards non aces)=n(all five cards nonaces)n(any five cards)
=48!5!43!52!5!47!
=48!5!43!5!47!52!
=47!48!43!52!
=43!4445464748!43!48!49505152

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-02-21

We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 192 had kids. Based on this, plot a 99% confidence interval for the proportion of adult residents who are parents in a given county.
Express your answer in the form of three inequalities. Give your answers in decimal fractions up to three places <p< Express the same answer using a point estimate and a margin of error. Give your answers as decimals, to three places.
p=±

asked 2021-03-18

A population of values has a normal distribution with μ=133.5 and σ=5.2. You intend to draw a random sample of size n=230.
Find the probability that a single randomly selected value is between 133.6 and 134.1.
P(133.6<X<134.1)=?
Write your answers as numbers accurate to 4 decimal places.

asked 2022-01-19
How is the law of large numbers related to probability?
asked 2021-05-30
Random variables X and Y have joint PDF
fX,Y(x,y)={12e(3x+4y), x0,y00, otherwise
Find P[max(X,Y)0.5]
asked 2020-11-16

Let XN(6,4).Find the probabilities P(5<X<7).

asked 2022-01-18
What are the mean and standard deviation of a binomial probability distribution with n=25 and p=1017?
asked 2022-03-13
If the joint probability density function of two continuous random variables X and Y isgiven by f(x; y) = 2, 0 < y < 3x, 0 < x < 1;find,(a) f(y/x), (3)(b) E(Y/x), (2)(c) Var(Y/x).