Recent content by MrGremlin

http://www.mathhelpboards.com/members/i-like-serena/, Thank you very much.:) So, b) P(A) - no face cards. there are 12 face cards in the deck.(including aces) 30-12 = 20 cards. $P(A)= \dfrac {C^{20}_3} {4960}$ P(B)= the same number of colors. $P(\clubsuit + \diamondsuit+ \heartsuit+...

P(E)-without spades and no aces 32-4-6=22 $P(E)=\dfrac {C^{20}_3} {4960} = 1140/4960$

Well. I try to explain my solution. 1) P(C) - [FONT=arial]at least one ace there are 32 cards in my deck. 4 of them are aces. 32-4=28. P(D)-there are no aces. $P(D)=\dfrac{C^ {28}_3} {4960}$ P(C) = 4960 - P(D) Where is my mistake?

is it all wrong?

P(C) = [FONT=arial]at least one ace.= $1-\dfrac{ C^{28}_3} {N}= (4960-3276)/4960 = 1684/4960$ I try to add the probabilities myself.[FONT=arial] b) P(A) - without face cards. $P(A)= \dfrac{C^{20}_3} {4960} = 1140/4960$ P(B) - i don't understand. We have different number of suits. So i should...

Hearts: 6,7,8,9,10,Queen,King,Ace Spades:6,7,8,9,10,Queen,Ace Diamonds:6,7,8,9,10.Jack,Queen,King,Ace Clubs:6,7,8,9,10.Jack,King,Ace

1)N=C^3_32=4960 P(A)- there are only spades, P(B) - there are no spades. My deck without King of spades and Jack of spades. So I have only 7 spades. P(A) = ((C^0_4)*(C^3_29))/4960=3654/4960 P(B)= 1-(3654/4960) = 1306/4960 Right? I doubt it.

the deck consist of 36 cards. And 4 of them are selected and removed - - - Updated - - - thank you for this video, but my problem isn't solved

Three cards are randomly selected, without replacement, from a deck of 32. (without King of spades, Jack of hearts, jack of spades and queen of clubs) Find the probability a) That the cards are without spades and of choosing at least one ace b) of choosing the same suit and there are no face...