A nickel is tossed 20 times in succession. Every time the nickel comes up heads, a dime is tossed, Let X count the number of heads appearing on tosses of the dime. Determine P(X=0) and P(X=1).
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To determine P(X=0) and P(X=1), we need to find the probability of obtaining zero heads on tosses of the dime (X=0) and the probability of obtaining exactly one head on tosses of the dime (X=1).
Let's break down the problem step by step:
Step 1: Determine the probability of getting no heads on the dime (X=0):
Since the dime is only tossed when the nickel comes up heads, to get zero heads on the dime, we need to get tails on all 20 nickel tosses.
The probability of getting tails on a single toss of a fair coin is 1/2. Therefore, the probability of getting tails on all 20 tosses is (1/2)^20.
Step 2: Determine the probability of getting exactly one head on the dime (X=1):
To get exactly one head on the dime, we need to have only one heads on the nickel in 20 tosses, and then get heads on the dime.
The probability of getting heads on a single toss of a fair coin is 1/2. Therefore, the probability of getting exactly one head on the nickel in 20 tosses is given by the binomial probability formula:
P(X=k) = (n choose k) * p^k * q^(n-k)
Where n is the number of trials (number of tosses), k is the number of successful outcomes (number of heads), p is the probability of success on a single trial, and q is the probability of failure on a single trial.
In our case, n = 20, k = 1, p = 1/2, and q = 1 - p = 1/2.
Plugging these values into the formula, we get:
P(X=1) = (20 choose 1) * (1/2)^1 * (1/2)^(20-1)
Simplifying this expression, we find:
P(X=1) = 20 * (1/2)^20
Now we have calculated both P(X=0) and P(X=1).