The probability of a head occurring when a biased coin is tossed is 0.58. If the coin is tossed six times, what is the probability of obtaining at most two heads in the six tosses??
binomial probability
(t + h)^6 ... (.42 + .58)^6 ... the 1st three terms of the expansion
p = .42^6 + 6(.42^5 * .58) + 15(.42^4 * .58^2) ...
To find the probability of obtaining at most two heads in six tosses, we can use the binomial probability formula.
The binomial probability formula is given by:
P(x) = (nCx) * (p^x) * ((1-p)^(n-x))
Where:
P(x) is the probability of getting exactly x successes.
n is the total number of trials (in this case, the number of coin tosses).
x is the number of desired successes (in this case, the number of heads).
p is the probability of a single success (in this case, the probability of getting a head).
nCx is the binomial coefficient, which represents the number of ways to choose x items from a set of n items.
In this problem:
p = 0.58 (probability of getting a head)
n = 6 (number of coin tosses)
We need to calculate the probability of at most two heads, which means we need to find the probabilities of getting 0, 1, and 2 heads, and then add them together.
Probability of getting 0 heads (x = 0):
P(0) = (6C0) * (0.58^0) * ((1-0.58)^(6-0))
Probability of getting 1 head (x = 1):
P(1) = (6C1) * (0.58^1) * ((1-0.58)^(6-1))
Probability of getting 2 heads (x = 2):
P(2) = (6C2) * (0.58^2) * ((1-0.58)^(6-2))
To calculate (nCx), we can use the formula:
nCx = n! / (x! * (n-x)!)
Using the above formulas, we can calculate the probabilities and then add them together.
P(0) = (6! / (0! * (6-0)!)) * (0.58^0) * ((1-0.58)^(6-0))
P(1) = (6! / (1! * (6-1)!)) * (0.58^1) * ((1-0.58)^(6-1))
P(2) = (6! / (2! * (6-2)!)) * (0.58^2) * ((1-0.58)^(6-2))
Once we calculate these probabilities, we can add them together to find the probability of obtaining at most two heads in six tosses.
To find the probability of obtaining at most two heads in six tosses of a biased coin with a probability of 0.58 for a head, we will calculate the individual probabilities for obtaining 0, 1, and 2 heads, and then sum them up.
Step 1: Calculate the probability of obtaining 0 heads:
The probability of obtaining a tail (opposite of a head) is 1 - 0.58 = 0.42.
Since we want to find the probability of obtaining 0 heads, we raise the probability of getting a tail to the power of 6 (the number of tosses).
P(0 heads) = (0.42)^6
Step 2: Calculate the probability of obtaining 1 head:
We multiply the probability of obtaining a head (0.58) by the probability of obtaining a tail (0.42) for the remaining 5 tosses. This can occur in 6 different ways (H-T-T-T-T, T-H-T-T-T, T-T-H-T-T, T-T-T-H-T, T-T-T-T-H).
P(1 head) = 6 * (0.58)^1 * (0.42)^5
Step 3: Calculate the probability of obtaining 2 heads:
We multiply the probability of obtaining a head (0.58) by the probability of obtaining a tail (0.42) for the remaining 4 tosses. This can occur in 15 different ways (H-H-T-T-T, H-T-H-T-T, H-T-T-H-T, H-T-T-T-H, T-H-H-T-T, T-H-T-H-T, T-H-T-T-H, T-T-H-H-T, T-T-H-T-H, T-T-T-H-H, H-T-H-H-T, H-H-H-T-T, H-H-T-T-H, H-T-T-H-H, T-T-H-H-H).
P(2 heads) = 15 * (0.58)^2 * (0.42)^4
Step 4: Sum up the probabilities from steps 1, 2, and 3 to find the probability of obtaining at most two heads:
P(at most 2 heads) = P(0 heads) + P(1 head) + P(2 heads)
P(at most 2 heads) = (0.42)^6 + 6 * (0.58)^1 * (0.42)^5 + 15 * (0.58)^2 * (0.42)^4