A coin is tossed 13 times. How many different outcomes have at least 2 heads ?
To find the number of outcomes with at least 2 heads when a coin is tossed 13 times, we can use the concept of combinations.
First, let's find the number of outcomes with exactly 2 heads. Since there are 13 tosses and exactly 2 of them are heads, we need to choose 2 out of 13 tosses to be heads. This can be calculated using the formula for combinations: C(n, r) = n! / (r!(n-r)!), where n is the total number of items and r is the number of items we want to choose.
Using this formula, we can calculate the number of combinations for 13 tosses with exactly 2 heads as C(13, 2) = 13! / (2!(13-2)!) = 78.
Next, let's consider the number of outcomes with exactly 3 heads. Following the same logic as before, we need to choose 3 out of 13 tosses to be heads. So, the number of combinations for 13 tosses with exactly 3 heads is C(13, 3) = 13! / (3!(13-3)!) = 286.
We can continue this process for each number of heads from 2 to 13. Therefore, the number of outcomes with at least 2 heads would be the sum of the number of outcomes with exactly 2 heads, exactly 3 heads, and so on until exactly 13 heads.
Summing up all the combinations, we get:
C(13, 2) + C(13, 3) + C(13, 4) + C(13, 5) + C(13, 6) + C(13, 7) + C(13, 8) + C(13, 9) + C(13, 10) + C(13, 11) + C(13, 12) + C(13, 13).
Evaluating this expression gives us the total number of different outcomes with at least 2 heads when a coin is tossed 13 times.