What is the probability of being dealt four aces in a five-card poker hand?

[six789]

i have no idea which technique of probability to choose. here is the problem...
1. a five-card poker hand is dealt and the last card is turned face up. Determine the probability that you have been dealt four aces, given that the card turned over is an ace.

2. I how many ways can a group of 10 people be chosen from 6 adults and 8 children if the group must contain at least 2 adults

i tried #1, i get like (52 choose 5)*(5 choose 4) but the book turns out to be 1/20825
and for the second one, i have no clue where to start

NOTE: can anyone help me to find a way when to use the different kinds of technique in probability, coz I am so confused...

 

[sruthisupriya]

try this

2. I how many ways can a group of 10 people be chosen from 6 adults and 8 children if the group must contain at least 2 adults
..

try no:adults=x and u want p(x>=2).summation 6Cx*8C(10-x)/14C10.start with x=2.

 

[Tide]

#1 HINT: You already know one of the cards leaving only 51 to choose from. In how many ways can you select four of the remaining cards such that one of them is not an ace?

 

[six789]

tide, i still cannot get the right answer. like i solve many times, but i really can't find the correct answer... i tried this one... 5C4*51*1/52C4, but its is wrong since the right answer in my book is 1/20825

 

[six789]

anyways, i found the solution for #2

but still i don't get the question in #1

 

[Tide]

The number of ways of selecting 3 aces and 1 card (not an ace) from the remaining 51 cards is C(3,3)XC(48, 1). The total number of ways of selecting any 4 cards from the remaining 51 given that one ace is already revealed is C(4,1)XC(51, 4) since the revealed card could be anyone of the 4 original aces.

 

[six789]

tide, ur so good! with ur explanation, i really understand it well...thanks man!