Mathematics, 17.01.2020 06:31 nessabear9472

Abridge hand is found by taking 13 cards at random and without replacement from a deck of 52 play- ing cards. find the probability of drawing each of the following hands. (a) one in which there are 5 spades, 4 hearts, 3 diamonds, and 1 club. (b) one in which there are 5 spades, 4 hearts, 2 diamonds, and 2 clubs. (c) one in which there are 5 spades, 4 hearts, 1 diamond, and 3 clubs. (d) suppose you are dealt 5 cards of one suit, 4 cards of another. would the probability of having the other suits split 3 and 1 be greater than the probability of having them split 2 and 2?

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Answer from: nataliamontirl8152

Part a

The probability of drawing each of one in which there are 5 spades, 4 hearts, 3 diamonds, and 1 clubs is 0.0054.

Part b

The probability of drawing each of one in which there are 5 spades, 4 hearts, 2 diamonds, and 2 clubs is 0.0088.

Part c

The probability of drawing each of one in which there are 5 spades, 4 hearts, 3 diamonds, and 1 clubs is 0.0054.

Part d

The probability of having the other suits split 3 and 1 is not greater than the probability of having them split 2 and 2. That is, \left( {0.0159 < 0.1350} \right)(0.0159<0.1350).

Step-by-step explanation:

Fundamentals

Let n be the total number of observations and r is the number of success then the combination is defined as,

$^n{C_r} = \frac{{n!}}{{\left( {n - r} \right)!r!}} The probability of an event is defined as,

[tex]\begin{array}{c}\\{\rm{Probability}} = \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{favorable}}\,{\rm{outcomes}}\,{\rm{for}}\,{\rm{an}}\,{\rm{event}}}}{{{\rm{Total}}\,{\rm{number}}\,{\rm{of}}\,{\rm{outcomes}}}}\\\\ = \frac{{N\left( {{\rm{event}}} \right)}}{{N\left( S \right)}}\\\end{array}$

(a)

The probability of drawing each of one in which there are 5 spades, 4 hearts, 3 diamonds, and 1 clubs is obtained below:

There are 52 playing cards in a deck out of that at randomly taking 13 cards without replacement. Let event X denotes the set of all passwords.

The total number of possible ways that 13 cards are drawing from deck of 52 cards is,

$N\left( S \right){ = ^{52}}{C_{13}}$

The number of possible ways that 13 cards are drawing each with 5 spades, 4 hearts, 3 diamonds, and 1 clubs is,

$N\left( X \right){ = ^{13}}{C_5}{ \times ^{13}}{C_4}{ \times ^{13}}{C_3}{ \times ^{13}}{C_1}$

The required probability is,

$\begin{array}{c}\\{\rm{Probability}} = \frac{{N\left( X \right)}}{{N\left( S \right)}}\\\\ = \frac{{^{13}{C_5}{ \times ^{13}}{C_4}{ \times ^{13}}{C_3}{ \times ^{13}}{C_1}}}{{^{52}{C_{13}}}}\\\\ = 0.0054\\\end{array}$

The probability of drawing each of one in which there are 5 spades, 4 hearts, 3 diamonds, and 1 clubs is 0.0054.

(b)

The probability of drawing each of one in which there are 5 spades, 4 hears, 2 diamonds, and 2 clubs is obtained below:

The total number of possible ways that 13 cards are drawing from deck of 52 cards is,

$N\left( S \right){ = ^{52}}{C_{13}}$

The number of possible ways that 13 cards are drawing each with 5 spades, 4 hears, 2 diamonds, and 2 clubs,

$N\left( X \right){ = ^{13}}{C_5}{ \times ^{13}}{C_4}{ \times ^{13}}{C_2}{ \times ^{13}}{C_2}$

The required probability is,

$\begin{array}{c}\\{\rm{Probability}} = \frac{{N\left( X \right)}}{{N\left( S \right)}}\\\\ = \frac{{^{13}{C_5}{ \times ^{13}}{C_4}{ \times ^{13}}{C_2}{ \times ^{13}}{C_2}}}{{^{52}{C_{13}}}}\\\\ = 0.0088\\\end{array}$

The probability of drawing each of one in which there are 5 spades, 4 hearts, 2 diamonds, and 2 clubs is 0.0088.

The probability of drawing each of one in which there are 5 spades, 4 hearts, 1 diamonds, and 3 clubs is obtained below:

The total number of possible ways that 13 cards are drawing from deck of 52 cards is,

$N\left( S \right){ = ^{52}}{C_{13}}$

The number of possible ways that 13 cards are drawing each with 5 spades, 4 hears, 1 diamonds, and 3 clubs is,

$N\left( X \right){ = ^{13}}{C_5}{ \times ^{13}}{C_4}{ \times ^{13}}{C_1}{ \times ^{13}}{C_3}$

The required probability is,

$\begin{array}{c}\\{\rm{Probability}} = \frac{{N\left( X \right)}}{{N\left( S \right)}}\\\\ = \frac{{^{13}{C_5}{ \times ^{13}}{C_4}{ \times ^{13}}{C_1}{ \times ^{13}}{C_3}}}{{^{52}{C_{13}}}}\\\\ = 0.0054\\\end{array}$

The probability of drawing each of one in which there are 5 spades, 4 hearts, 3 diamonds, and 1 clubs is 0.0054.

(d)

The probability of having the other suits split 3 and 1 be greater than the probability of having them split 2 and 2 is obtained below:

There are 5 cards of one suit, 4 cards of another.

A suit consists of 13 cards out of which 5 are drawn this implies that there are 8 cards remains in the suit $8\left( { = 13 - 5} \right)8(=13−5)[tex]. After removing 13 cards from the deck, the remaining cards are [tex]39\left( { = 52 - 13} \right)39(=52−13)$ out of which 4 are drawn this implies that 35 cards remains $35\left( { = 39 - 4} \right)35(=39−4)$ .

The probability of having the other suits split 3 and 1 is,

$\begin{array}{c}\\{\rm{Probability}} = \frac{{\left( {^8{C_3}{ \times ^{35}}{C_1}} \right)}}{{\left( {^{43}{C_4}} \right)}}\\\\ = 0.0159\\\end{array}$

The probability of having the other suits split 2 and 2 is,

$\begin{array}{c}\\{\rm{Probability}} = \frac{{\left( {^8{C_2}{ \times ^{35}}{C_2}} \right)}}{{\left( {^{43}{C_4}} \right)}}\\\\ = 0.1350\\\end{array}$

Answer from: Quest

)because it's going in order of the math sign

Answer from: Quest

step-by-step explanation:

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