please help probabilities?

Question

box A contains 5 white balls and 7 red balls and box b contains 6 white and 4 red balls. A box is chosen at random and then a ball is drawn from it. the probability that the ball is red is
4/11
1/2
59/120
4

Answer 1

You have a 50% (0.5) chance of picking box A.
You have a 50% (0.5) chance of picking box B.

If you pick box A, you have a 7/(5+7) chance of getting a red ball.
If you pick box B, you have a 4/(6+4) chance of getting a red ball.

So your answer is
=(1/2)(7/12) + (1/2)(4/10)
=7/24 + 4/20
=35/120 + 24/120
=59/120 or ~49.17%

Answer 2

Use The Law of Total Probability

For a set of events A1, A2, A3, ... , An where the Ai's are mutually exclusive and exhaustive events and for any other event B

P(B)
= P(B and A1) + P(B and A2) + ... + P(B and An)
= P(B | A1) * P(A1) + P(B | A2) * P(A2) + ... + P(B | An) * P(An)

here you have:

P( red | box A ) = 7/12
P( red | box B ) = 4/10

P(box A) = P(box B) = 0.50

P(red) = P(red | box A) * P(box A) + P(red | box B) * P(box B)
P(red) = 7/12 * 1/2 + 4/10 * 1/2
P(red) = 59/120 = 0.4916667

Answer 3

= 0.5*(7/12) + 0.5*(4/10) = 7/24 + 1/5 = 59/120 = 0.491666667

Answer 4

I THINK you're supposed to calculate the probability for each box and then add them...

P(red ball) = (1/2 * 7/12) + (1/2 * 4/10) = (7/24) + (4/20) = 59/120 = 49%

Answer 5

P(R | A) = 7/12
P(R | B) = 4/10

P(R) = 7/12 * 1/2 + 4/10 * 1/2
= 59/120

NOTE that 11/22 = 1/2 is a trick.