ask for help about maths - probability(10 marks)

First of all, let's summarize all possible events that will lead to drawing a red ball from bag Y finally:

1) Not drawing a red ball from bag X but a red ball from bag Y:

P(Not drawing a red ball from bag X) = 1/4

Then after putting the ball into bag Y, bag Y will have a total of 10 balls with 6 being red, thus:
P(Drawing a red ball from bag Y) = 6/10 = 3/5

Using the multiplication of probability, the probability of event 1 is:
(1/4) x (3/5) = 3/20

2) Drawing a red ball from bag X and then a red ball from bag Y:

P(Not drawing a red ball from bag X) = 3/4

Then after putting the ball into bag Y, bag Y will have a total of 10 balls with 7 being red, thus:
P(Drawing a red ball from bag Y) = 7/10

Using the multiplication of probability, the probability of event 1 is:
(3/4) x (7/10) = 21/40

Since events (1) and (2) are mutually exclusive (as you cannot say that in the first draw, you can draw A RED AND NOT A RED ball)

So P(Event 1 or 2) = P(Event 1) + P(Event 2)
= 3/20 + 21/40
= 27/40

So the answer is D.

Let us call the ball that drawn from bag x and put into bag y be Ball Q.

Now, Bag Y has 3 yellow balls, 6 red balls and a ball Q.Total 10 balls.

The probability of drawing a red ball is 6/10 = 3/5
The probability of drawing ball Q is 1/10
While the chance of ball Q that will be a red ball is 3/4
therefore the probability of drawing ball Q which is a red ball is 1/10 x 3/4 = 3/40
therefore, the total probability of drawing a red ball from bag Y is
3/5 + 3/40 = 27/40

So the answer is D.