The probability is 1 if you draw three balls without replacement.
If only one draw, it is 3/5.
17 out of 21
The probability that you pick 3 white balls is 0.05 or 5%.----------------------------------------------------------------------------------------------------EXPLANATIONThe probability of first drawing a white ball is: P(W1) =3/6.The probability of drawing a white ball given the event that you already draw awhite ball in the first draw and not replacing it back is: P(W2│W1) =2/5.The probability of drawing a white ball on the third draw given the event that awhite ball was drawn in the first and in the second draw is: P(W3│(W1UW2)) =1/4Now, the probability of drawing 3 white balls one by one with out replacement(taking all 3 balls at a time gets the same analysis and result) is:P(W1UW2UW3) =P(W1)∙P(W2│W1)∙P(W3│(W1UW2)) =(3/6)∙(2/5)∙(1/4) =1/20=0.05 =5%.
The probability of drawing a white marble is .46
It is (4/8)*(6/8) = 3/8
14/27 ~ 0.519 ~ 51.9%----------------------------------------------------------------------------------------------------EXPLANATIONFor bag 1: Probability of drawing a black ball; P(B1) =6/9. A white one; P(W1) =3/9For bag 2: P(B2) =5/9. P(W2) =4/9.Probability of drawing a black ball from each bag: P(B1UB2) =(6/9)∙(5/9) =10/27Probability of drawing a white ball from each bag: P(W1UW2) =(3/9)∙(4/9) =4/27The probability of drawing 2 balls of the same color is the sum of the aboveprobabilities:P(2 same color balls) =P(B1UB2) + P(W1UW2) =10/27 +4/27 =14/27 =0.5185...P(2 same color balls) ~ 0.519 ~ 51.9%
In case I haven't explained that very well here's an example: I have a bag containing black and white balls and I think that one quarter of the balls are black. I start drawing balls from the bag (and putting them back afterwards) and I keep drawing out white balls. If I were continue to pick only white balls how many times would this have to happen before I was 90% sure that less than ¼ balls were black? Thanks
The probability of getting 3 white balls in a draw of 5 balls with replacement from an urn containing white balls and black balls is always greater than the same test without replacement, because the number of white balls decreases when you draw a white ball and do not replace it. The ratio of white to black with replacement is constant, and is always less than one, assuming there is at least one black ball. The ratio of white to black without replacement decreases each turn, and is still less than one, and is less than the previous ratio, unless the question asked about 2 white balls or less.
2/7 = 28.57%
17 out of 21
The probability that you pick 3 white balls is 0.05 or 5%.----------------------------------------------------------------------------------------------------EXPLANATIONThe probability of first drawing a white ball is: P(W1) =3/6.The probability of drawing a white ball given the event that you already draw awhite ball in the first draw and not replacing it back is: P(W2│W1) =2/5.The probability of drawing a white ball on the third draw given the event that awhite ball was drawn in the first and in the second draw is: P(W3│(W1UW2)) =1/4Now, the probability of drawing 3 white balls one by one with out replacement(taking all 3 balls at a time gets the same analysis and result) is:P(W1UW2UW3) =P(W1)∙P(W2│W1)∙P(W3│(W1UW2)) =(3/6)∙(2/5)∙(1/4) =1/20=0.05 =5%.
The probability is zero, because there are no red balls in the bag.
The probability of drawing a white marble is .46
The probability of a simple event is the number of ways to succeed over the total possibilities. So, there are three white balls, and ten balls total. So: P(white) = 3/10
It is (4/8)*(6/8) = 3/8
The question cannot be answered because it is incomplete and a complete version cannot be found.
14/27 ~ 0.519 ~ 51.9%----------------------------------------------------------------------------------------------------EXPLANATIONFor bag 1: Probability of drawing a black ball; P(B1) =6/9. A white one; P(W1) =3/9For bag 2: P(B2) =5/9. P(W2) =4/9.Probability of drawing a black ball from each bag: P(B1UB2) =(6/9)∙(5/9) =10/27Probability of drawing a white ball from each bag: P(W1UW2) =(3/9)∙(4/9) =4/27The probability of drawing 2 balls of the same color is the sum of the aboveprobabilities:P(2 same color balls) =P(B1UB2) + P(W1UW2) =10/27 +4/27 =14/27 =0.5185...P(2 same color balls) ~ 0.519 ~ 51.9%
A drawing done in only one color. a black and white drawing is considered monochrome, as the "color" would be black on a white tapestry.