A bag contains n blue and m red marbles. You randomly pick a marble from the bag, write down its color, and then put the marble back in the bag. This process is repeated until you pick either two consecutive blue or two consecutive red marbles. Given that the process stopped because you picked two consecutive blue marbles, what is the probability that the first marbled you picked was blue

(mn+n²)/(m+n)

Step-by-step explanation:

probability of blue marble= n/(n+m)

probability of red marble= m/(n+m)

probability that process stops = Probability of both blue + probability of both red=  n/(n+m) × n/(n+m) + m/(n+m)×m/(n+m)

                        = (n²+m²)/(n+m)²

P(1st marbel is blue)= P(blue and red) + P(blue and blue)

                                  = mn/(n+m) + n²/(n+m)

                                   = (mn+n²)/(m+n)

P(1st marble is blue | process stops)= ( (mn+n²)/(m+n)× (n²+m²)/(n+m)²)/ ((n²+m²)/(n+m)²)

= (mn+n²)/(m+n)

0

step-by-step explanation:

you can not divide by 0 so 4 ÷ 0 would be 0

0 × 9999 =0

mark brainliest   : )

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632