A coin is tossed 7 times. What is the probability that the number of heads obtained will be between 2 and 4 inclusive? Express your answer as a fraction or a decimal number rounded to four decimal places.
To find the probability of obtaining a certain number of heads when tossing a coin, we need to understand that a fair coin has two equally likely outcomes: heads (H) or tails (T). Thus, the probability of getting heads or tails on a single toss is both 1/2.
Now, we need to find the probability of getting between 2 and 4 inclusive heads when tossing the coin 7 times.
To solve this problem, we can use the binomial probability formula, which states that the probability of getting exactly k successes in n independent trials, each with the same probability of success (p), is given by:
P(k) = (n choose k) * p^k * (1 - p)^(n - k),
where (n choose k) is the binomial coefficient, equal to n! / (k! * (n-k)!).
In our case, n = 7 (the number of coin tosses), p = 1/2 (the probability of heads), and we want to find P(2) + P(3) + P(4).
Let's calculate each term:
P(2) = (7 choose 2) * (1/2)^2 * (1 - 1/2)^(7 - 2) = (21) * (1/4) * (1/2)^5 = 21/128.
P(3) = (7 choose 3) * (1/2)^3 * (1 - 1/2)^(7 - 3) = (35) * (1/8) * (1/2)^4 = 35/256.
P(4) = (7 choose 4) * (1/2)^4 * (1 - 1/2)^(7 - 4) = (35) * (1/16) * (1/2)^3 = 35/512.
Now, let's calculate the sum:
P(2-4) = P(2) + P(3) + P(4) = 21/128 + 35/256 + 35/512 ≈ 0.3096.
Thus, the probability of obtaining between 2 and 4 inclusive heads when tossing a coin 7 times is approximately 0.3096, rounded to four decimal places.