PLEASE HELP- ALGEBRA/PROBABILITY QUESTION:

5 white balls and k black balls are placed into a bin. Two of the balls are drawn at random. The probability that one of the drawn balls in white and the other is black is 10/21. Find the smallest possible value of k.

Question

PLEASE HELP- ALGEBRA/PROBABILITY QUESTION:

5 white balls and k black balls are  placed into a bin. Two of the balls are drawn at random. The probability that one of the drawn balls in white and the other is black is 10/21. Find the smallest possible value of k.

uri · Answer

@terenzreignz Is bored,He will help you.

ybarrap · Answer

solve - $$5k/((5+k(4+k) + 5k/((5+k)(5+k-1) = 10/21$$
you will get k=2 or k=10
So smallest k is 2

anonymous · Answer

Thanks, I believe that is correct

ybarrap · Answer

The two terms of the sum are probability of getting white black or black white

anonymous · Answer

Cause 5k= choosing 1 white and 1 black, then I *think* you can find the total number of choosing two by doing 7c2.

anonymous · Answer

So 5k= 10, and 7 c 2 = 21

ybarrap · Answer

Equation should be $$5k/((5+k)(4+k)) + 5k/((5+k)(5+k-1)) = 10/21$$

ybarrap · Answer

$$(1/2)*5k/((5+k)(4+k))=10/21 $$

ybarrap · Answer

The rest is algebra, quadratic equation stuff

ybarrap · Answer

That last equation should have been:
$$2∗5k/((5+k)(4+k))=10/21 $$

ybarrap · Answer

Because both P(wb) and P(bw) are equal

ybarrap · Answer

All the algebraic steps are here:

Solve for k over the real numbers:
(10 k)/((k+4) (k+5)) = 10/21
Multiply both sides by a polynomial to clear fractions.
Cross multiply:
210 k = 10 (k+4) (k+5)
Write the quadratic polynomial on the right hand side in standard form.
Expand out terms of the right hand side:
210 k = 10 k^2+90 k+200
Move everything to the left hand side.
Subtract 10 k^2+90 k+200 from both sides:
-10 k^2+120 k-200 = 0
Factor the left hand side.
The left hand side factors into a product with three terms:
-10 (k-10) (k-2) = 0
Divide both sides by a constant to simplify the equation.
Divide both sides by -10:
(k-10) (k-2) = 0
Solve each term in the product separately.
Split into two equations:
k-10 = 0 or k-2 = 0
Look at the first equation:  Solve for k.
Add 10 to both sides:
k = 10 or k-2 = 0
Look at the second equation:  Solve for k.
Add 2 to both sides:
Answer: |  
 | k = 10 or k = 2

ybarrap · Answer

Remember, you are drawing a ball without replacement. So the the second term (representing the second ball drawn) in either P(wb) or P(bw) will have one less ball in the denominator

ybarrap · Answer

Does this make sense?