A BOY CONTAINS RED AND BLUE MARBLES ,TWO MARBLES ARE DRAWN WIHT OUT REPLACEMENT.THE PROBABLILITY OF SELECTING A RED MARBLE ANDTHEN A BLUE MARBLE IS 0.28,THE PROBABILITY OF SELECTING A RED MARBLE ON THE FIRST DRAWNIS 0.5, WHAT IS THEPROBABILITY OF SELCTING A BLUE MARBLE ON THE SECOND DRAW,GIVEN THE FIRST MARBLE DRAWN WAS RED?
r/(b+r) * b/(b+r-1) = .28
.5 * b/(b+r-1) = .28
b/(b+r-1) = .56
See where that takes you, given what you know about conditional probabilities.
To find the probability of selecting a blue marble on the second draw, given that the first marble drawn was red, we can use the concept of conditional probability. Let's break down the information given:
The probability of selecting a red marble on the first draw is 0.5.
The probability of selecting a red marble and then a blue marble without replacement is 0.28.
To find the probability of selecting a blue marble on the second draw, given that the first marble drawn was red, we need to use the conditional probability formula:
P(B|A) = P(A and B) / P(A)
Where:
P(B|A) is the probability of event B occurring given that event A has already occurred.
P(A and B) is the probability of both A and B occurring.
P(A) is the probability of event A occurring.
In this case, event A is drawing a red marble on the first draw, and event B is drawing a blue marble on the second draw. Therefore, we need to find P(B|A).
The probability of drawing a red marble on the first draw is 0.5. So, P(A) = 0.5.
The probability of drawing a red marble and then a blue marble without replacement is 0.28. So, P(A and B) = 0.28.
Using the formula for conditional probability, we can now calculate P(B|A):
P(B|A) = P(A and B) / P(A)
P(B|A) = 0.28 / 0.5
P(B|A) = 0.56
Therefore, the probability of selecting a blue marble on the second draw, given that the first marble drawn was red, is 0.56 or 56%.