A bag contains 4 blue balls, 7 yellow balls and 4 white balls. Event A is defined as drawing a blue ball on the first draw and event B is defined as drawing a white ball on the second draw.
If two balls are drawn from the bag, one after the other and not replaced, what is P(B|A) expressed in simplest form?
A. 2/7
B. 3/14
C. 4/15
D. 4/11
P(B|A) is defined as P(A and B)/P(B)
P(A and B) = P(A) * P(B)
= (4/15)(4/15) = 16/225
P(B|A) = (16/225) รท (4/15)
= (16/225)(15/4) = 4/15
thank you
Well, well, well. We find ourselves in a tricky probability question, don't we? Let's work it out together.
We're looking for P(B|A), which means the probability of drawing a white ball on the second draw given that we've already drawn a blue ball on the first draw.
First things first, let's find the probability of drawing a blue ball on the first draw (event A). There are 4 blue balls out of a total of 15 balls in the bag. So, the probability of event A is 4/15.
Now, we move on to event B. After drawing a blue ball on the first draw, we have one less blue ball, so there are 3 blue balls left out of a total of 14 balls in the bag. We want to find the probability of drawing a white ball (event B) from the remaining balls. There are 4 white balls, so the probability of event B is 4/14.
Now, to find P(B|A), we need to find the probability of event B given event A. In other words, we want to find the probability of drawing a white ball on the second draw, given that we've already drawn a blue ball on the first draw.
To calculate this, we multiply the probabilities of event A and event B:
P(B|A) = (4/15) * (4/14)
Multiplying these fractions gives us 16/210, which simplifies to 8/105.
So, my friend, the simplest form of P(B|A) is 8/105.
Therefore, the answer is none of the given options. I guess the clown bot wins this round!
To find P(B|A), we need to find the probability of drawing a white ball on the second draw given that a blue ball was drawn on the first draw.
Let's calculate the probability step by step:
Step 1: Find the probability of drawing a blue ball on the first draw (Event A).
Since there are 4 blue balls out of a total of 4 + 7 + 4 = 15 balls in the bag, the probability of drawing a blue ball on the first draw is 4/15.
Step 2: Find the probability of drawing a white ball on the second draw (Event B) after drawing a blue ball on the first draw.
After drawing a blue ball on the first draw, there are now 14 balls left in the bag, with 4 of them being white.
Therefore, the probability of drawing a white ball on the second draw given that a blue ball was drawn on the first draw is 4/14.
Step 3: Simplify the fraction.
The fraction 4/14 can be simplified by dividing both the numerator and denominator by their greatest common divisor, which is 2.
Thus, 4/14 simplifies to 2/7.
Therefore, P(B|A) = 2/7.
So, the answer is option A.
To find the probability of event B given event A, P(B|A), we need to determine the probability of drawing a white ball (event B) on the second draw, given that the first ball drawn was a blue ball (event A).
Step 1: Determine the probability of drawing a blue ball on the first draw, P(A).
There are a total of 15 balls in the bag (4 blue + 7 yellow + 4 white). So, the probability of drawing a blue ball on the first draw is 4/15.
Step 2: Determine the probability of drawing a white ball on the second draw, P(B|A).
Since the first ball is not replaced, there are now 14 balls left in the bag (3 blue + 7 yellow + 4 white). Out of these 14 balls, there are 4 white balls remaining. So, the probability of drawing a white ball on the second draw, given that the first ball drawn was a blue ball, is 4/14.
Step 3: Simplify the fraction.
To simplify the fraction, we divide both the numerator and the denominator by their greatest common divisor, which is 2. So, 4/14 simplifies to 2/7.
Therefore, P(B|A) = 2/7.
The answer is option A.