- #1
F.B
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Sorry to post so many questions.
But i need help.
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.
I think its a conditional probability problem but here's what i did instead.
Probability of the first to be an ace is: 4/52
Second card: 3/51
Third: 2/50
Fourth to not be an ace: 48/49
Fifth:1
I multiply all those together but i don't get the right answer.
2. In how many ways can a group of ten people be chosen from 6 adults and 8 children if the group must contain at least 2 adults.
I did this.
C(14,10) - C(6,0) x C(8,10) - C(6,1) x C(8,9)
This doesn't work out for some reason. I thinks how you suppose to do it but it won't work.
3. There are 25 students in a data management class. Determine the probability that at least two share the same birthday. (Assume that every year has 365 days)
This one i have no clue how to do it.
But i need help.
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.
I think its a conditional probability problem but here's what i did instead.
Probability of the first to be an ace is: 4/52
Second card: 3/51
Third: 2/50
Fourth to not be an ace: 48/49
Fifth:1
I multiply all those together but i don't get the right answer.
2. In how many ways can a group of ten people be chosen from 6 adults and 8 children if the group must contain at least 2 adults.
I did this.
C(14,10) - C(6,0) x C(8,10) - C(6,1) x C(8,9)
This doesn't work out for some reason. I thinks how you suppose to do it but it won't work.
3. There are 25 students in a data management class. Determine the probability that at least two share the same birthday. (Assume that every year has 365 days)
This one i have no clue how to do it.