SOLUTION: Two balls are drawn in succession out of a box containing 5 red and 3 white balls. Find the probability that at least 1 ball was red, given that the first ball was
(A) Replaced
Algebra.Com
Question 1138273: Two balls are drawn in succession out of a box containing 5 red and 3 white balls. Find the probability that at least 1 ball was red, given that the first ball was
(A) Replaced before the second draw.
(B) Not replaced before the second draw.
I tried utilizing the formula P(AnB)=P(A)P(BIA)=P(B)P(AIB)
I have no clue how to solve with the given variable of a different number of white balls.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Part A
Define two events
A = drawing a white ball on the first selection
B = drawing a white ball on the second selection
P(A) = 3/8 since there are three white balls out of 3+5 = 8 total.
There is replacement, so P(B|A) = P(B) = P(A), which is possible because of the independent events and because we're focused on the same color for each event.
We can then say,
P(A and B) = P(A)*P(B|A)
P(A and B) = (3/8)*(3/8)
P(A and B) = 9/64
Now onto the main probability we want to find. We subtract the last answer from 1. This is because the two events of "at least one red" and "no reds" are complementary in nature.
P(At least one red) = 1 - P(no reds)
P(At least one red) = 1 - P(two white)
P(At least one red) = 1 - P(A and B)
P(At least one red) = 1 - 9/64
P(At least one red) = 64/64 - 9/64
P(At least one red) = (64 - 9)/64
P(At least one red) = 55/64
Final Answer for part A: 55/64
This answer is exact in terms of a fraction.
If you need the answer in decimal form, then use your calculator to find that 55/64 = 0.859375 which converts to 85.9375%
side note: The formula you provided
P(A and B) = P(A)*P(B|A)
can be changed into
P(A and B) = P(A)*P(B)
but can only be done if the events are independent. This is because P(B|A) = P(B) when the events are independent.
===========================================================================================
Part B
Again we define the following
A = drawing a white ball on the first selection
B = drawing a white ball on the second selection
This time however, P(B|A) = P(B) is not true due to the fact that replacement is not made. The event A happening changes the probability P(B). Normally P(B) = 3/8 if replacement is made, but now P(B) = 2/7 as there are 3-1 = 2 white balls left out of 8-1 = 7 total. Therefore, P(B|A) = 2/7 if replacements aren't made. We say that these events are dependent. Specifically B depends on A. Saying "A depends on B" is not true.
So,
P(A and B) = P(A)*P(B|A)
P(A and B) = (3/8)*(2/7)
P(A and B) = 6/56
P(A and B) = 3/28
and,
P(At least one red) = 1 - P(no reds)
P(At least one red) = 1 - P(two white)
P(At least one red) = 1 - P(A and B)
P(At least one red) = 1 - 3/28
P(At least one red) = 28/28 - 3/28
P(At least one red) = (28 - 3)/28
P(At least one red) = 25/28
Final Answer for part B: 25/28 (exact answer fraction form)
The decimal form is approximately 25/28 = 0.892857 which converts to 89.2857%
RELATED QUESTIONS
a box contains 3 red, 3 white, and 3 green balls. Two balls are drawn out of the box in... (answered by stanbon)
A box contains 4 red, 3 white and 4 green balls. Two balls are drawn out of the... (answered by stanbon,greenestamps)
A box contains 4 red, 2 white and 3 green balls. Two balls are drawn out of the box in (answered by Alan3354,ikleyn)
A box contains 2 red, 4 white and 4 green balls. Two balls are drawn out of the... (answered by Boreal)
A box contains 44 red, 33 white and 44 green balls. Two balls are drawn out of the (answered by Shin123,ikleyn)
A box contains 2 red, 4 white and 2 green balls. Two balls are drawn out of the box in (answered by rothauserc)
Three boxes containing red, white and blue balls are used in an experiment. Box 1... (answered by CPhill)
two balls are drawn at random without replacement from a box containing 6 red balls, 4... (answered by ewatrrr)
A box contains 12 red balls and 8 white balls. Two balls are drawn in succession. Find... (answered by flame8855)