A fair coin is tossed 10 times. What is the probability of getting exactly 6 heads?
by the method of binomial theorem,
P = C(n,r).p^r.q^n-r
= C(10,6).(1/2)^6.(1/2)^4
= 210/1024
= 105/512 Ans.
Correct! Good job.
To find the probability of getting exactly 6 heads when a fair coin is tossed 10 times, we first need to determine the total number of possible outcomes and the number of favorable outcomes.
When a coin is tossed, there are two possible outcomes, heads (H) or tails (T). Since there are two options for each toss, the total number of possible outcomes for 10 coin tosses is 2 raised to the power of 10, which equals 1024.
To calculate the number of favorable outcomes, we need to determine the number of ways to get exactly 6 heads in 10 tosses. This can be done using the binomial coefficient, which represents the number of possible combinations.
The binomial coefficient can be calculated using the formula: C(n, k) = n! / (k! * (n-k)!)
In this case, we want to find C(10, 6) because there are 10 coin tosses and we want exactly 6 heads. Plugging these values into the formula, we get:
C(10, 6) = 10! / (6! * (10-6)!) = 10! / (6! * 4!)
= (10 * 9 * 8 * 7 * 6 * 5 * 4!) / (6! * 4!)
= (10 * 9 * 8 * 7 * 6 * 5) / (6 * 5 * 4 * 3 * 2 * 1)
= 10 * 9 * 8 * 7 = 5040
Therefore, there are 5040 favorable outcomes for exactly 6 heads in 10 tosses.
Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 5040 / 1024
Probability ≈ 0.0491
Therefore, the probability of getting exactly 6 heads when a fair coin is tossed 10 times is approximately 0.0491 or 4.91%.