Probability question drawing cards with replacement?

P(at least 7 clubs)=P( 7 clubs from 19)+P(8 clubs from 19)+...+P(19 clubs from 19)

In a deck of cards, there are 13 clubs. therefore, P(club)=13/52=1/4 and

P(no club)=1-13/52=39/52=3/4

P(7 clubs)=P(7 clubs and (19-7=8) not clubs)=P(7 clubs)*P(12 not clubs)

=(13/52)^7 * (39/52)^12

but there are many ways to get 7 clubs from the 19 chosen, namely

19C7=19!/[7!*(19-7)!]=50388 ways

therefore, P(7 clubs)=(19C7) * (13/52)^7 * (39/52)^12 = 0.09742 (approx.)

note the general formula P(x clubs)=(19Cx) * (13/52)^x * (39/52)^(19-x)

do the same for x=8,9,...,19 and add the probabilities together to get the required probability.

an alternative way

P(X=>7) = 1-P(X<=6) by complementary probabilities

= 1-[P(X=0)+P(X=1)+...+P(X=6)]

P(at least 7 clubs)=1-P(less than and including 6 clubs)

the final answer should be 0.1749 (approx.)

ie. 17.49% chance to get 7 or more clubs if a card is chosen at random with replacement

note: [P(X=0)+P(X=1)+...+P(X=6)] + [P(X=7)+...+P(X=19)]=1

[P(X=7)+...+P(X=19)] = 1- [P(X=0)+P(X=1)+...+P(X=6)]

Chance of getting one club per draw = 13 / 52 = 1 / 4

Chance of getting no club = 1 - 1 / 4 = 3 / 4

So chance of getting at least one club in 19 draws with

replacement = 1 - (3 / 4)^19

So chance of getting at least 7 clubs = (1 - (3 / 4)^19)^7 = 0.971

For the reason that without alternative now we have for a) there are 13 clubs in a deck of playing cards so, hazard is 13 clubs/52 cards x 12 golf equipment/fifty one cards consequently, 13/52 x 12/51=1/4 x four/17=four/68 keep in mind thirteen golf equipment then 12 clubs b/c possibility of drawing one membership for denominator, start with fifty two cards then one drawn, so, 51 playing cards next b) each red. Identical notion but 26 crimson playing cards in a deck. So 26/fifty two x 25/51=half x 25/fifty one=25/102 c) equal suit?? So, there are four suits all having 13 playing cards. As a consequence, 4 approaches of opting for a suit and 13C4 approaches of settling on four playing cards from a given suit. The 13C4 is from permutation, combinations, factorials. We receive the 13 from 13 playing cards of each go well with and the 4 from having 4 suits in a deck here is where i'm hoping that i dont lose you four x 13C4/52C4 the place n=thirteen r=four in n umerator the place n=fifty two r=4 in denominator utilizing n!/(n-r)!R! Thirteen!/(thirteen-4)!4!/fifty two!/(52-four)!Four!=572/54145 hope this helps!! If you would like more of a proof write to me here and i'll supply a more specific rationalization!!