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A fair die with faces {1, 2, 3, 4, 5, 6} is thrown repeatedly till ‘3’ is observed for the first time. Let X denote the number of times the die is thrown. The expected value of X is _____.
Correct answer is '6'. Can you explain this answer?
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A fair die with faces {1, 2, 3, 4, 5, 6} is thrown repeatedly till &ls...
given
now, the random variable X does represent number of throws required for getting 3. So,
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A fair die with faces {1, 2, 3, 4, 5, 6} is thrown repeatedly till &ls...
Till a 6 is obtained. Let X be the number of throws needed. We are asked to find the expected value of X.
To solve this problem, we can use the concept of geometric distribution. The geometric distribution models the number of trials needed to obtain a success (in this case, rolling a 6) in a sequence of independent trials.
The probability of rolling a 6 on any single throw of the die is 1/6, since the die is fair. Therefore, the probability of rolling a 6 on the first throw is 1/6.
The probability of not rolling a 6 on the first throw is 5/6. In this case, we start a new sequence of trials and repeat the process until a 6 is obtained. The probability of rolling a 6 on the second throw (given that a 6 was not rolled on the first throw) is again 1/6.
Using this reasoning, we can see that the probability of rolling a 6 on the nth throw (given that a 6 was not rolled on any of the previous n-1 throws) is also 1/6.
Therefore, we can model the number of throws needed to obtain a 6 as a geometric distribution with probability of success p = 1/6.
The expected value (or mean) of a geometric distribution with success probability p is given by E(X) = 1/p. In this case, we have p = 1/6, so the expected value of X is E(X) = 1/(1/6) = 6.
Therefore, the expected value of the number of throws needed to obtain a 6 is 6.
To solve this problem, we can use the concept of geometric distribution. The geometric distribution models the number of trials needed to obtain a success (in this case, rolling a 6) in a sequence of independent trials.
The probability of rolling a 6 on any single throw of the die is 1/6, since the die is fair. Therefore, the probability of rolling a 6 on the first throw is 1/6.
The probability of not rolling a 6 on the first throw is 5/6. In this case, we start a new sequence of trials and repeat the process until a 6 is obtained. The probability of rolling a 6 on the second throw (given that a 6 was not rolled on the first throw) is again 1/6.
Using this reasoning, we can see that the probability of rolling a 6 on the nth throw (given that a 6 was not rolled on any of the previous n-1 throws) is also 1/6.
Therefore, we can model the number of throws needed to obtain a 6 as a geometric distribution with probability of success p = 1/6.
The expected value (or mean) of a geometric distribution with success probability p is given by E(X) = 1/p. In this case, we have p = 1/6, so the expected value of X is E(X) = 1/(1/6) = 6.
Therefore, the expected value of the number of throws needed to obtain a 6 is 6.
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A fair die with faces {1, 2, 3, 4, 5, 6} is thrown repeatedly till ‘3’ is observed for the first time. Let X denote the number of times the die is thrown. The expected value of X is _____.Correct answer is '6'. Can you explain this answer?
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