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A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is [2002]
- a)8/3
- b)3/8
- c)4/5
- d)5/4
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A dice is tossed 5 times. Getting an odd number is considered a succes...
The event follows binomial distribution with
n = 5, p = 3/6 = 1/2. q = 1 – p = 1/2.;
∴ Variance = npq = 5/4.
n = 5, p = 3/6 = 1/2. q = 1 – p = 1/2.;
∴ Variance = npq = 5/4.
Most Upvoted Answer
A dice is tossed 5 times. Getting an odd number is considered a succes...
Solution:
Given that, getting an odd number is considered a success.
Let X be the random variable defined as the number of successes in 5 tosses.
We know that the probability of getting an odd number on a single toss is 1/2.
Therefore, the probability of getting a success on a single toss is 1/2.
Now, we can use the binomial distribution to find the probability distribution of X.
The probability mass function (pmf) of X is given by:
P(X=k) = (5Ck)(1/2)^k(1/2)^(5-k) for k=0,1,2,3,4,5
where 5Ck is the binomial coefficient.
Using the formula for variance of a binomial distribution, we get:
Var(X) = np(1-p)
where n=5 is the number of trials and p=1/2 is the probability of success on a single trial.
Substituting the values, we get:
Var(X) = 5(1/2)(1/2) = 5/4
Therefore, the variance of the distribution of successes is 5/4.
Hence, the correct option is (D) 5/4.
Given that, getting an odd number is considered a success.
Let X be the random variable defined as the number of successes in 5 tosses.
We know that the probability of getting an odd number on a single toss is 1/2.
Therefore, the probability of getting a success on a single toss is 1/2.
Now, we can use the binomial distribution to find the probability distribution of X.
The probability mass function (pmf) of X is given by:
P(X=k) = (5Ck)(1/2)^k(1/2)^(5-k) for k=0,1,2,3,4,5
where 5Ck is the binomial coefficient.
Using the formula for variance of a binomial distribution, we get:
Var(X) = np(1-p)
where n=5 is the number of trials and p=1/2 is the probability of success on a single trial.
Substituting the values, we get:
Var(X) = 5(1/2)(1/2) = 5/4
Therefore, the variance of the distribution of successes is 5/4.
Hence, the correct option is (D) 5/4.
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A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is [2002]a)8/3b)3/8c)4/5d)5/4Correct answer is option 'D'. Can you explain this answer?
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