A coin is tossed 6 times. what is the probability of getting at least two heads?
To find the probability of getting at least two heads when a coin is tossed 6 times, we need to consider the different possible outcomes.
First, let's calculate the probability of getting exactly two heads in 6 tosses:
The number of ways to get exactly two heads in 6 tosses can be calculated using the combination formula: C(n, r) = n! / (r! * (n-r)!), where n is the total number of tosses and r is the number of heads.
In this case, n = 6 (the total number of tosses) and r = 2 (the number of heads we want).
C(6, 2) = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
The total number of possible outcomes for 6 coin tosses is 2^6 = 64 (since each toss has 2 possibilities: head or tail).
Therefore, the probability of getting exactly two heads in 6 tosses is 15/64.
Next, we need to consider the probability of getting three heads, four heads, five heads, and six heads. Following the same steps as before, we can calculate the probabilities for each of these cases:
- Getting exactly three heads: C(6, 3) = 6! / (3! * (6-3)!) = 20
- Getting exactly four heads: C(6, 4) = 6! / (4! * (6-4)!) = 15
- Getting exactly five heads: C(6, 5) = 6! / (5! * (6-5)!) = 6
- Getting exactly six heads: C(6, 6) = 6! / (6! * (6-6)!) = 1
Now, we add up the probabilities of each of these cases to find the probability of getting at least two heads:
P(at least two heads) = P(exactly two heads) + P(exactly three heads) + P(exactly four heads) + P(exactly five heads) + P(exactly six heads)
= 15/64 + 20/64 + 15/64 + 6/64 + 1/64
= 57/64
Therefore, the probability of getting at least two heads when a coin is tossed 6 times is 57/64.