Compute E(X) for the following random variable X :
X=Number of tosses until getting 4 (including the last toss) by tossing a fair 10-sided die.
To compute E(X) for the given random variable X, which represents the number of tosses until getting a 4 by tossing a fair 10-sided die, we need to calculate the mathematical expectation or the expected value.
The expected value of a discrete random variable can be found by multiplying each possible value of the variable by its corresponding probability and summing up all these values.
In this case, let's analyze the problem step by step:
Step 1: Identify the possible values of X.
The possible values of X are the positive integers from 1 to infinity, as it takes at least one toss to obtain a 4.
Step 2: Calculate the probability of obtaining a 4 on each toss.
Since the 10-sided die is fair, each outcome (numbers 1 to 10) has an equal probability of 1/10.
Step 3: Calculate the probability of needing a certain number of tosses to get a 4.
To obtain a 4 on the first toss, the probability is 1/10.
To obtain a 4 on the second toss, the probability is (9/10) * (1/10).
To obtain a 4 on the third toss, the probability is (9/10) * (9/10) * (1/10).
And so on...
Step 4: Calculate E(X) using the formula:
E(X) = ∑(k * P(X=k)), where k is the value of the random variable X and P(X=k) is the probability of X being equal to k.
For this particular random variable, it becomes:
E(X) = 1/10 + 2 * ((9/10) * (1/10)) + 3 * ((9/10) * (9/10) * (1/10)) + ...
Now, we can calculate E(X) by summing up the series, which represents the expected number of tosses until getting a 4.
To compute the expected value of X, we need to determine the probability distribution of X and then calculate the sum of X multiplied by their respective probabilities.
In this case, we can start by considering the possible values of X. Since X represents the number of tosses until getting a 4, it can take on values from 1 to infinity.
Let's calculate the probability distribution of X step-by-step:
1. The probability of getting a 4 on the first toss is 1/10, since the die has 10 equally likely outcomes.
P(X=1) = 1/10
2. The probability of getting a 4 on the second toss is given by the probability of not getting a 4 on the first toss (9/10) times the probability of getting a 4 on the second toss (1/10).
P(X=2) = (9/10) * (1/10) = 9/100
3. Similarly, the probability of getting a 4 on the third toss is given by the probability of not getting a 4 on the first two tosses (9/10)^2 times the probability of getting a 4 on the third toss (1/10).
P(X=3) = (9/10)^2 * (1/10) = 81/1000
4. Following the same pattern, the probability of getting a 4 on the fourth toss is (9/10)^3 * (1/10).
P(X=4) = (9/10)^3 * (1/10) = 729/10000
Continuing this pattern, we can see that the probability of getting a 4 on the n-th toss is given by:
P(X=n) = (9/10)^(n-1) * (1/10)
Now that we have the probability distribution of X, we can compute the expected value:
E(X) = Σ (X * P(X)) for all possible values of X
∞
E(X) = Σ (n * P(X))
n=1
Substituting the values of P(X), we get:
E(X) = 1/10 + 2 * (9/100) + 3 * (81/1000) + 4 * (729/10000) + ...
To compute the exact value of E(X), we would need to calculate the infinite sum. However, we can see that the sum converges to a finite value, as each term in the sum gets smaller and smaller.
In this case, we can use the formula for the sum of an infinite geometric series:
Sum = a / (1 - r)
where a is the first term and r is the ratio between consecutive terms.
In this case, the first term is 1/10 and the ratio is 9/10. Plugging these values into the formula, we get:
E(X) = (1/10) / (1 - 9/10)
= (1/10) / (1/10)
= 1
Therefore, the expected value of X is 1.