A box contains some white balls and some blue balls. There are 5 more blue balls than white balls. One ball is removed at random and not replaced. A second ball is then removed at random. The probability that the balls are different colours is 52/105. Find the probability that both balls are white. Show your working out.
8 answers
16/25
robert can you explain that pease
Make an equation,
(W x B) + (B x W) = 52/105
This can be simplified to a quadratic equation, the positive value for x is 8;
then replace this value of x in (W x W), the answer is 2/15.
(W x B) + (B x W) = 52/105
This can be simplified to a quadratic equation, the positive value for x is 8;
then replace this value of x in (W x W), the answer is 2/15.
I can't take a photo of my working out, but continuing from Robaesa's work,
x= white balls
x+5=blue balls
2x+5 is the total in first draw, 2x+4 is the total in 2nd draw
Pr(WB) x Pr(BW) = 52/105
Using algebra your final quadratic equation should be
x^2+57x-520=0
x=8 or -65
you have 8 white balls, and now you can answer the question easily
x= white balls
x+5=blue balls
2x+5 is the total in first draw, 2x+4 is the total in 2nd draw
Pr(WB) x Pr(BW) = 52/105
Using algebra your final quadratic equation should be
x^2+57x-520=0
x=8 or -65
you have 8 white balls, and now you can answer the question easily
I don’t get it lol
I am sorry but I don't understand your answers.
If i have 13 blue balls what does the quadratic equation fraction multiplied by the constant K did on the vector AB = b + 2??
I'm sorry, but your question doesn't seem to be related to the previous problem. Could you please provide more information or clarify your question?
1 answer