A bag contains 10 counters six of them are white account was taken at random and not replaced second kind was taken out of the bag at random calculate the probability that only one of the two counters is white

Correct question is;

A bag contains 10 counters. 6 of them are white. A counter is taken at random and not replaced. A second counter is taken out of the bag at random. Calculate the probability that only one of the two counters are white

probability that only one of the two counters is white = 8/15

Step-by-step explanation:

To solve this question, first of all, let's look at probability we would have to either draw two white counters or two non-white counters (4/10 * 3/9).

Probability(draw 2 white counters) = (6/10 × 5/9) = 30/90 = 1/3

Probability(draw 2 non-white counters) = (4/10 × 3/9) = 2/15

Now, In all other cases, we'll draw exactly one white and one non-white counter, so the odds of this would be;

P(one white counter and one non-white counter) = 1 - [1/3 + 2/15)

= 1 - 7/15 = 8/15

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: