A bag contains 3 green balls. 2 red balls and a white ball. A boy randomly draws balls from the bag one at a time (with replacement) until a red ball appears. Find the probability that he will make at least 3 draws.
He could get it on
the first draw
or the 2nd draw
or the 3rd draw
or the 4th draw
or the 5th draw
or the 6th draw --- , at this point he must have picked it
= 2/6 + (4/6)(2/5) + (4/6)(3/5)(2/4) + (4/6)(3/5)(2/4)(2/3) + (4/6)(3/5)(2/4)(1/3)(2/2) + 0
= 1/3 + 4/15 + 1/5 + 2/15 + 1/15
notice that this adds up to 1
so prob of at least 3 draws = 1 - (1/3 + 4/15) = 2/5
make a tree chart. one scenario would be drawing a red ball, another a white ball, another...etc. Then, after you pick one case, make other choices... like after i picked a red ball...i picked either a red ball again or a white ball...
but Reiny it's WITH replacement!
I got 4/5 but the answer in the book says something else. Am I correct?
Sorry I meant 5/9
Right! , should read the question more carefully
so my string would be
(2/6) + (4/6)(2/6) + (4/6)(4/6)(2/6) + (4/6)(4/6)(4/6)(2/6) + .....
= (1/3) + (2/3)(1/3) + (2/3)^2 (1/3) + (2/3)^3 (1/3) + ...
This must have a total of 1
so the prob of least 3 draws
= 1 - (1/3 + 2/9) = 4/9
note my string of additions is a geometric series
where a = 1/3 and r = 2/3
sum(∞) = a/(1-r)
= (1/3) / (1-2/3) = 1
To find the probability that the boy will make at least 3 draws until a red ball appears, we need to consider the different possible scenarios.
First, let's calculate the probability of not drawing a red ball in the first 2 draws.
The probability of not drawing a red ball on the first draw is given by:
P(not red on first draw) = (number of non-red balls) / (total number of balls) = 4/6 = 2/3
Now, since the draws are made with replacement, the probability of not drawing a red ball on the second draw is the same as the probability of not drawing a red ball on the first draw:
P(not red on second draw) = 2/3
Using the multiplication rule for independent events, we multiply the probabilities of the individual draws to find the probability of not drawing a red ball in the first 2 draws:
P(not red in first 2 draws) = P(not red on first draw) * P(not red on second draw) = (2/3) * (2/3) = 4/9
Next, let's calculate the probability of drawing a red ball on the third or subsequent draws. Since the previous draws were not red, the remaining balls in the bag are: 3 green balls, 1 red ball, and 1 white ball (a total of 5 balls).
The probability of drawing a red ball on the third or subsequent draws is given by:
P(red on third or subsequent draw) = (number of red balls) / (total number of balls) = 1/5
Finally, we need to calculate the complement of not drawing a red ball in the first 2 draws, which is obtaining a red ball within the first 2 draws.
The probability of drawing a red ball within the first 2 draws is given by:
P(drawing red in first 2 draws) = 1 - P(not red in first 2 draws) = 1 - 4/9 = 5/9
Therefore, the probability that the boy will make at least 3 draws until a red ball appears is:
P(at least 3 draws) = P(red on third or subsequent draw) = 1/5