A balanced coin is tossed 8 times. What is the probability of getting exactly 2 heads given that there were at least 2 heads in the 8 tosses?
To find the probability of getting exactly 2 heads given that there were at least 2 heads in the 8 tosses, we need to use conditional probability. Let's break down the problem into two parts:
Part 1: Calculate the probability of getting at least 2 heads in 8 tosses.
To calculate this, we need to consider all the possible outcomes. In each coin toss, there are two possible outcomes - heads or tails. So, for 8 tosses, we have 2^8 = 256 possible outcomes.
Now, let's determine the number of outcomes where we have at least 2 heads. We can use the complement rule to simplify the calculation. The complement of "at least 2 heads" is "0 or 1 head." So we need to subtract the number of outcomes with 0 or 1 head from the total number of outcomes.
The number of outcomes with 0 or 1 head can be calculated with the binomial coefficient. The binomial coefficient, denoted as "C(n, k)," represents the number of ways to choose k elements from a set of n elements without regard to their order. The formula for the binomial coefficient is C(n, k) = n! / (k!(n-k)!).
In this case, we want to calculate C(8, 0) and C(8, 1):
C(8, 0) = 8! / (0!(8-0)!) = 1
C(8, 1) = 8! / (1!(8-1)!) = 8
The number of outcomes with at least 2 heads is:
Total number of outcomes - Number of outcomes with 0 or 1 head
= 256 - (C(8, 0) + C(8, 1))
= 256 - (1 + 8)
= 256 - 9
= 247
So, the probability of getting at least 2 heads in 8 tosses is 247/256.
Part 2: Calculate the probability of getting exactly 2 heads given that there were at least 2 heads.
Since we know that there were at least 2 heads in the 8 tosses, we can consider the reduced sample space of 247 outcomes. Now, we need to calculate the number of outcomes with exactly 2 heads and divide it by the total number of outcomes in this reduced sample space.
To calculate the number of outcomes with exactly 2 heads, we can use the binomial coefficient again:
C(8, 2) = 8! / (2!(8-2)!) = 28
So, the probability of getting exactly 2 heads given that there were at least 2 heads in the 8 tosses is 28/247.
In summary, the probability of getting exactly 2 heads given that there were at least 2 heads in the 8 tosses is 28/247.