(1 point) All that is left in a packet of candy are 7 reds, 3 greens, and 3 blues.

(a)What is the probability that a random drawing yields a red followed by a green assuming that the first candy drawn is put back into the packet?

(b)Are the events 'red' and 'green' dependent?


I GOT 10/26 1/13 AND IT SAYS THE ANSWER IS WRONG?

A 21 /169

B independent

Step-by-step explanation:

The total number of candies is 7+3+3 = 13

P( 1st red) = red/total = 7/13

Since the candy is put back there are still 13 candies in the dish

P(2nd green) = green/total = 3/13

P( red, then green) = P( 1st red) * P(2nd green)

                                = 7/13* 3/13 = 21/169

The events are independent since the candies are put back

1     sepan

2     encuentren

3     tenga

4     hable

5   parezca

6   llegue