Probability problem...?

Question

A bag contains four balls, three are red and the fourth is either red or blue, selected with equal prbability. A ball is randomly drawn from the bad, and it is red. What is the probability that one of the balls still in the bag is blue?

Answer 1

If there were two bags, one with all red and one with 3 red, one blue, and you plci a bag at random then pick a marble from the bag, there are 8 possible outcomes. 7 of them are pick red. Discard the other one. 3 of those 7 came from the bag with the blue marble, so the probability that one of the balls in the bag is still blue is 3/7.

Answer 2

You are looking for the conditional probability that the one ball in question is blue given that a red ball was drawn from the bag.
Pr(B|R) = Pr(B&R)/Pr(R) = 3/7

Here Pr(B&R) is the probability that a red ball was drawn from a bag containing 3 red balls and 1 blue and is equal to
(1/2) x (3/4) = 3/8
Pr(R) is the probability to draw a red ball in both scenarios and is equal to (1/2) x (3/4) + (1/2) x 1 = 7/8

Answer 3

there are 3 balls left. The chance of the ball being blue is 1/2 out of the two colors and the chance of getting that ball is 1/3 so it is 1/6

Answer 4

Two cases: RRRR or RRRB.
If the first case were true, the probability of drawing a red was 1.
If the second case were true, the probability of drawing a red was 3/4.
Thus P(second case given you drew a red) = P(second case and you drew a red)/P(you drew a red) = 3/4 / (1 + 3/4) = 3/7.

Answer 5

i favor the ten undesirable--SO i presumed this one outONEOUT AND MY answer IS.........27% danger OF GETTIN THE ALL HEARTS IN sequence FROM AN ACE to 10. that's HOW I ARRIVED at my determination, ok???? pay interest properly reason i purely madeit to point 3 after too lengthy of a time and now i'm attempting element 4. if i had fifty 2 playing cards that are in a deck of playing cards and u dealt me 5 like u suggested u did--that would want to go away me with 40 seven playing cards. minus a 10 would equivalent 37p.c.and the percentages on that are even worse-r than that so upload yet another or sub-music yet another 10 for the percentages bein hostile to me and that i purely ended up with a 27% and am i ceremony or rong??? i'm a kia

Answer 6

The people who say 3/7 are correct.
I think that stephen m's explanation is clearest.

Answer 7

let a: bag contains 3 r & 1b
b: bag contains 4r
e: ball taken out is red
we have to find P(a/e):probability of a given e
USING BAYE'S THEOREM WE HAVE
p(a/e)=p(e/a)p(a)/(p(e/a)p(a)+p(e/b)p(b))
p(e/a)=3/4
p(e/b)=1
p(a)=p(b)=1/2
therefore
p(e/a)=3/7

Answer 8

i think the answer will be 1/3