Probability closed

Question

A bag contain 3R and 5 B balls and second bag contain 6R and 4B  balls . A ball is drawn from each bag . find the probability that both are (i)red and (2) black.?

Comments

According to me 1 and 2 number doesnt specify that first ball is red and second ball is black .

Let E be an event of getting a  red  ball

Let F be event of getting a black ball

Drawing a ball from one bag may not change the sample space of other bag . so they are independents events

P(E intersection F) = (3/8)(4/10) + (5/8)*(6/10 )

3/8 getting a red ball from first bag

4/10 getting a black ball from second bag

5/8 getting a red ball from second bag

6/10 getting a red ball from second bag

is it ryt ?

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one ball is drawn from each bag that means there are two balls and the questions if the both balls are red then it becomes 3/8+6/10 because For drawing a red ball from first back the probability becomes 3/8 and from second bag the probability becomes 6/10. Similarly the second case will be.
And in the case of second both are black then the probability becomes 5/8+4/10