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 im so confused...
#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?
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
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 any one of the 4 original aces.