Calvin and Dan are playing a game of chance. Calvin tosses 8 fair coins, and wins if he obtains at least 4 heads. The probability that Calvin wins can be expressed as a/b where a,b are coprime positive integers. What is the value of
a+b?
Use the symmetry w.r.t. interchanging heads and tails.
P(4 or more heads) = P(4 or more tails) = P(less than 5 heads).
So:
2 P(4 or more heads) = P(4 or more heads) + P(less than 4 heads) +
P(4 heads) = 1 + P(4 heads) ------>
P(4 or more heads) =
1/2 + 1/2 P(4 heads) =
1/2 + 1/2^9 Binomial(8,4) =
1/2 + 1/2^9 8*7*6*5/(4*3*2) =
1/2 + 35/2^8 =
(2^7 + 35)/2^8 =
163/256
a+b=419
To solve this problem, we need to determine the probability that Calvin wins the game by obtaining at least 4 heads out of the 8 coins tossed.
First, let's calculate the total number of possible outcomes when tossing 8 coins. Each coin has 2 possible outcomes - either a head or a tail. Since there are 8 coins, the total number of outcomes is 2^8 = 256.
Next, let's determine the number of favorable outcomes for Calvin winning the game. We need to consider the following cases:
1. Calvin gets exactly 4 heads: There are 8C4 = 70 ways to choose 4 coins out of 8 to be heads.
2. Calvin gets exactly 5 heads: There are 8C5 = 56 ways to choose 5 coins out of 8 to be heads.
3. Calvin gets exactly 6 heads: There are 8C6 = 28 ways to choose 6 coins out of 8 to be heads.
4. Calvin gets exactly 7 heads: There are 8C7 = 8 ways to choose 7 coins out of 8 to be heads.
5. Calvin gets exactly 8 heads: There is only 1 way to get all coins to be heads.
So, the total number of favorable outcomes is 70 + 56 + 28 + 8 + 1 = 163.
Therefore, the probability of Calvin winning the game is 163/256.
Since 163 and 256 share no common factors other than 1, the values of a and b are 163 and 256 respectively.
Finally, we can find the sum of a and b: a+b = 163+256 = 419.