a coin is tossed such that is twice as likely to show heads as tails. find the probability that in six tosses
a) exactly four heads obtained.
b) at most three heads are obtained.
To solve this problem, we first need to determine the probability of heads and tails for a single coin toss. Let's denote the probability of heads as P(H) and the probability of tails as P(T).
According to the given information, the coin is twice as likely to show heads as tails. This means that P(H) = 2 * P(T).
Since the sum of probabilities must equal 1, we have the equation: P(H) + P(T) = 1.
Substituting 2 * P(T) for P(H) in the equation, we can solve for P(T):
2 * P(T) + P(T) = 1
3 * P(T) = 1
P(T) = 1/3
Now we can determine P(H) using P(T):
P(H) = 2 * P(T)
P(H) = 2 * 1/3
P(H) = 2/3
Now that we know the individual probabilities for heads and tails, we can proceed to answer the question.
a) To find the probability of exactly four heads obtained in six tosses, we need to calculate the probability of getting four heads and two tails in any order.
The number of ways to arrange four heads and two tails in six tosses is given by the binomial coefficient, denoted as "n choose k," which is calculated as:
n! / (k! * (n - k)!)
In this case, n = 6 (number of tosses) and k = 4 (number of heads).
Plugging in the values, we get:
(6! / (4! * (6 - 4)!)) * (P(H)^4) * (P(T)^2)
Simplifying, we have:
(6! / (4! * 2!)) * (2/3)^4 * (1/3)^2
Calculating further:
(6 * 5 * 4! / (4! * 2!)) * (2/3)^4 * (1/3)^2
Simplifying:
(6 * 5) / (2 * 1) * (2/3)^4 * (1/3)^2
Which gives us:
15 * (2/3)^4 * (1/3)^2
Evaluating the expression:
15 * (16/81) * (1/9)
15 * 16/729
240/729 or approximately 0.3296
Therefore, the probability of obtaining exactly four heads in six tosses is approximately 0.3296.
b) To find the probability that at most three heads are obtained in six tosses, we need to calculate the probabilities of getting zero, one, two, and three heads, and then sum them up.
To calculate the probability of zero heads (all tails), we use:
P(T)^6 = (1/3)^6
To calculate the probability of one head, we need to consider all possible positions of the head among six tosses:
6 * P(H) * P(T)^5 = (6 * 2/3 * 1/3^5)
To calculate the probability of two heads, we use the binomial coefficient:
(6! / (2! * (6 - 2)!)) * (P(H)^2) * (P(T)^4) = (15 * (2/3)^2 * (1/3)^4)
To calculate the probability of three heads, we use the binomial coefficient:
(6! / (3! * (6 - 3)!)) * (P(H)^3) * (P(T)^3) = (20 * (2/3)^3 * (1/3)^3)
Now, sum up the probabilities for zero, one, two, and three heads:
(P(T)^6) + (6 * P(H) * P(T)^5) + (15 * (2/3)^2 * (1/3)^4) + (20 * (2/3)^3 * (1/3)^3)
(1/3)^6 + (6 * 2/3 * 1/3^5) + (15 * (2/3)^2 * (1/3)^4) + (20 * (2/3)^3 * (1/3)^3)
(1/729) + (12/729) + (60/729) + (160/729)
(233/729) or approximately 0.3191
Therefore, the probability that at most three heads are obtained in six tosses is approximately 0.3191.