A wallet contains 3 dimes, 6 pennies, and 6 nickels. Event A is defined as drawing a dime on the first draw, and event B is defined as drawing a dime on the second draw.
If Jack draws two coins from the wallet, one after the other without replacement, what is P(B|A) expressed in simplest form?
A.
B.
C.
D.
A. 1/9
B. 1/8
C. 2/9
D. 1/4
Just like most of your other posts dealing with conditional probability,
P(B|A) by definition
= P(B and A)/P(A)
now P(A) = 3/15 = 1/5
P(A and B) = ((1/5)(2/14) = ...
etc
thank you!
To find the probability of event B occurring given that event A has already occurred, we can use the concept of conditional probability. The probability of B given A is denoted as P(B|A) and can be calculated using the formula:
P(B|A) = P(A and B) / P(A)
To calculate P(A and B), we need to find the probability of drawing a dime on the first draw (event A) and then drawing a dime on the second draw (event B). Since the coins are drawn without replacement, the outcomes of the two events are dependent.
Let's start by calculating P(A), the probability of drawing a dime on the first draw. There are a total of 3 + 6 + 6 = 15 coins in the wallet. Out of these, 3 are dimes. Therefore,
P(A) = 3/15
P(A) = 1/5
Now let's calculate P(A and B), the probability of drawing a dime on both the first and second draw. After drawing a dime on the first draw, there are now 14 coins remaining in the wallet (since one dime has been removed). Out of these, 2 are dimes. Therefore,
P(A and B) = 2/14
P(A and B) = 1/7
Now we can calculate P(B|A) using the formula:
P(B|A) = P(A and B) / P(A)
P(B|A) = (1/7) / (1/5)
P(B|A) = (1/7) * (5/1)
P(B|A) = 5/7
So, the probability P(B|A) expressed in simplest form is 5/7.
Therefore, the answer is option B.