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.
robert can you explain that pease
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
To find the probability that both balls are white, we need to determine the number of white balls and the total number of balls.
Let's assume that the number of white balls in the box is 'x'. Therefore, the number of blue balls will be 'x + 5' according to the given information.
The probability that the first ball is white is 'x / (x + x + 5)', which simplifies to 'x / (2x + 5)'.
After removing one ball, the total number of balls becomes '2x + 4'. Therefore, the probability that the second ball is white is '(x - 1) / (2x + 4)'.
Since both balls need to be white, we multiply these probabilities together:
(x / (2x + 5)) * ((x - 1) / (2x + 4)) = (x(x - 1))/((2x + 5)(2x + 4))
According to the given information, the probability that the balls are different colors is 52/105. We can set up the equation:
2 * (x / (2x + 5)) * ((x + 5) / (2x + 4)) = 52/105
To solve this equation, we can cross-multiply and simplify:
2(x^2 + 5x) = (52/105) * (4x^2 + 18x + 20)
2x^2 + 10x = (4/105) * (4x^2 + 18x + 20)
Multiplying both sides by 105 to eliminate the denominator:
210x^2 + 1050x = 4(4x^2 + 18x + 20)
210x^2 + 1050x = 16x^2 + 72x + 80
Rearranging the equation to form a quadratic equation:
194x^2 - 978x + 80 = 0
Now we can solve this quadratic equation for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values a = 194, b = -978, and c = 80 into the quadratic formula:
x = (-(-978) ± √((-978)^2 - 4(194)(80))) / (2(194))
Simplifying further:
x = (978 ± √(956484 - 62720)) / 388
Calculating the discriminant (√(956484 - 62720)):
x = (978 ± √(893764)) / 388
x = (978 ± 944) / 388
This gives us two possible values for x:
x = (978 + 944) / 388 = 1922/388 = 4.9588 (approximately)
x = (978 - 944) / 388 = 34/388 = 0.0876 (approximately)
However, we can ignore the second result (0.0876) since it is not a whole number value. Therefore, the number of white balls, x, is 4.
Now, substituting x = 4 back into the original equation to find the probability that both balls are white:
(x(x - 1))/((2x + 5)(2x + 4)) = (4(4 - 1))/((2 * 4 + 5)(2 * 4 + 4))
= 3 / (13 * 12)
= 3 / 156
= 1 / 52
Therefore, the probability that both balls are white is 1/52.
16/25
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?
I don’t get it lol
I am sorry but I don't understand your answers.
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.