+1836 I have a bag with 5 pennies and 3 nickels. I draw coins out one at a time at random. What is the probability that I haven't removed all 3 nickels after 4 draws?
+129852
This equals ... 1 - P(that you have removed all three nickels after four draws)
Here's the probabilities associated with this sequence of draws ... N N N P =
(3/8)(2/7)(1/6)(5/5) =1 / 56
And since the penny can be chosen in any of four positions, we have four different possible sequences, each with the same probability. So, the probability of choosing all three nickels in any four draws is just C(4,1) (1/56) = 4/56 = 1/14
So the probability that you haven't drawn all three nickels in four draws is just......
1 - (1/14) = 13/14 = about 92.86%
And as Melody might say.....that's what I think......!!!!
+129852
This equals ... 1 - P(that you have removed all three nickels after four draws)
Here's the probabilities associated with this sequence of draws ... N N N P =
(3/8)(2/7)(1/6)(5/5) =1 / 56
And since the penny can be chosen in any of four positions, we have four different possible sequences, each with the same probability. So, the probability of choosing all three nickels in any four draws is just C(4,1) (1/56) = 4/56 = 1/14
So the probability that you haven't drawn all three nickels in four draws is just......
1 - (1/14) = 13/14 = about 92.86%
And as Melody might say.....that's what I think......!!!!
