Probability of drawing less than two yellow marbles?
December 19th, 2009 | by admin |A bag contains 8 red marbles, 8 blue marbles, 10 white marbles and 8 yellow marbles. You are asked to draw 5 marbles from the bag without replacement. What is the probability drawing less than two yellow marbles?
There are 8 yellow marbles and 26 which are not yellow, a total of 34.
Probability of drawing less than 2 yellows
= probability of drawing 0 OR 1 yellow
= probability of drawing 0 yellows + probability of drawing 1 yellow.
Probability of drawing 5 marbles which are not yellow:
P(0)= (26×25x24×23x22)/(34×33x32×31x30)
Probability of drawing 1 yellow followed by 4 which are not yellow:
P(1)= (8/34)x(26×25x24×23)/(33×32x31×30)
The yellow can be drawn on 1st, 2nd, 3rd, 4th or 5th turn, so this probability must be multiplied by 5.
So the total probability is
P(0,1) = (5×8+22)x(26×25x24×23)/(34×33x32×31x30) = 0.6662.
4 Responses to “Probability of drawing less than two yellow marbles?”
By miryana kina on Dec 19, 2009 | Reply
answer
= [8C1 * 26C4 + 8C0 * 26C5] / 34C5
= [8 * 14950 + 1 * 65780] / 278256
= 185380 / 278256
= 0.666221
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By Brittanyy(: on Dec 19, 2009 | Reply
do your own homework(:
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By Kristofer B on Dec 19, 2009 | Reply
8+8+8+10=34 less than 2 is one, so 8/34=23.5%
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By Barry G on Dec 19, 2009 | Reply
There are 8 yellow marbles and 26 which are not yellow, a total of 34.
Probability of drawing less than 2 yellows
= probability of drawing 0 OR 1 yellow
= probability of drawing 0 yellows + probability of drawing 1 yellow.
Probability of drawing 5 marbles which are not yellow:
P(0)= (26×25x24×23x22)/(34×33x32×31x30)
Probability of drawing 1 yellow followed by 4 which are not yellow:
P(1)= (8/34)x(26×25x24×23)/(33×32x31×30)
The yellow can be drawn on 1st, 2nd, 3rd, 4th or 5th turn, so this probability must be multiplied by 5.
So the total probability is
P(0,1) = (5×8+22)x(26×25x24×23)/(34×33x32×31x30) = 0.6662.
References :