Question
a bag contains 7 red chips and 10 blue chips. Two chips are selected randomly without replacement from the bag. what is the probability that both chips are red??
Answers
PsyDAG
Work similar to two black cards problem in previous post.
Twg
but how is the answer in decimals??
Marth
You can convert between fractions and decimals - the important part of this question is finding the probability.
Try finding the probability, in either format, and we will be happy to critique your work or help you convert answer formats.
Try finding the probability, in either format, and we will be happy to critique your work or help you convert answer formats.
Twg
I don't understand.
Marth
The total number of chips in the bag is 7+10 = 17.
The probability of drawing a red chip the first time is (# of red chips)/(total # chips) = _____________
The probability of drawing a red chip the second time, with no replacement, means that you should assume 1 red chip was removed the first time.
The probability of drawing a red chip the second time = (# of red chips -1)/(total # of chips) = ____________
Multiply the two probabilities together to get your final answer.
The probability of drawing a red chip the first time is (# of red chips)/(total # chips) = _____________
The probability of drawing a red chip the second time, with no replacement, means that you should assume 1 red chip was removed the first time.
The probability of drawing a red chip the second time = (# of red chips -1)/(total # of chips) = ____________
Multiply the two probabilities together to get your final answer.
Twg
I still do not get it.
I took (7-1)/17= and I got 0.352
but the answers that I have to choose from are
a) 0.485
b) 0.154
C) 0.515
D) 0.548
I am lost!!!
I took (7-1)/17= and I got 0.352
but the answers that I have to choose from are
a) 0.485
b) 0.154
C) 0.515
D) 0.548
I am lost!!!
PsyDAG
You need to use the <I>two</I> probabilities, 7/17 and (since one red has been removed) 6/16. The probability of <I>both</I> occurring is found by <I>multiplying</I> the probabilities of the individual events. Try it, and one answer will pop out for you.