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Mathematics 7 Online
OpenStudy (anonymous):

From a standard desk of 52 card, 5 card are draw.What are the odds of each even occurring? a) 5 aces b) 5 face cards

OpenStudy (ash2326):

a) 5 aces , there are 4 aces in a deck so probability of getting 5 aces is zero b) king queen and jack are face cards so total face cards= 12 no. of ways 5 face cards can be drawn= 12C5 no. of ways 5 cards can be drawn= 52C5 so probability of drawing 5 face cards= 12C5/ 52C5

OpenStudy (anonymous):

tk,

OpenStudy (ash2326):

welcome nancy did you understand

OpenStudy (anonymous):

I will give u a medal

OpenStudy (ash2326):

I didn't ask for that, I asked if you understand:)

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

actually it asks for odds, not probability, although the method of finding it is the same

OpenStudy (anonymous):

take the number of ways to get five face cards, divide by the number of ways not got to get five face cards

OpenStudy (anonymous):

number of ways to get 5 face cards is \[12\times 11\times 10\times 9\times 8=95040\] number of ways not to get five face cards is \[\dbinom{52}{5}-95040=2503920\]and so ODDS in favor of getting 5 face cards is \[95040:2503920\] or \[396:10433\]

OpenStudy (anonymous):

I have a question ash

OpenStudy (anonymous):

face have total 12 cards?

OpenStudy (ash2326):

King, queen and jack are face cards, so there are 3 in each suit so total 12

OpenStudy (anonymous):

I'm confuse

OpenStudy (ash2326):

Nancy In a deck of 52 cards, there are 4 suits each suit has = 1 queen, 1 king and 1 jack= 3 face cards so in 4 suits we'll have 12 cards

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