I assume you want exactly 4 heads of the 8 tosses
Prob = C(8,4) (1/2)^4 (1/2)^4
= ....
0.5. To the nearest thousandth, find the probability of getting 4 heads.
Prob = C(8,4) (1/2)^4 (1/2)^4
= ....
The binomial probability formula is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k),
where:
P(X = k) is the probability of getting exactly k successes,
C(n, k) is the number of combinations of n items taken k at a time (n choose k),
p is the probability of success on a single trial, and
n is the total number of trials.
In this case, k = 4 (the number of heads we want), p = 0.5 (the probability of heads on a single toss), and n = 8 (the total number of tosses).
To calculate C(n, k), we use the formula:
C(n, k) = n! / (k!(n - k)!)
Plugging in the values, we have:
C(8, 4) = 8! / (4!(8 - 4)!)
= 8! / (4! * 4!)
Calculating this expression gives us:
C(8, 4) = (8 * 7 * 6 * 5 * 4!) / (4! * 4 * 3 * 2 * 1)
= (8 * 7 * 6 * 5) / (4 * 3 * 2 * 1)
= 70
Now, we can plug this value into the binomial probability formula:
P(X = 4) = C(8, 4) * (0.5)^4 * (1 - 0.5)^(8 - 4)
P(X = 4) = 70 * 0.5^4 * 0.5^4
= 70 * (0.5)^8
Calculating this expression gives us:
P(X = 4) ≈ 0.273
Therefore, to the nearest thousandth, the probability of getting 4 heads when a coin is tossed 8 times is approximately 0.273.