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In 10 independent rollings of a biased die, the probability that an even number will appear 5 times is twice the probability that an even number will appear 4 times. What is the probability that an even number will appear twice when the die is rolled 8 times?
- a)0.0304
- b)0.1243
- c)0.2315
- d)0.1926
Correct answer is option 'A'. Can you explain this answer?
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In 10 independent rollings of a biased die, the probability that an ev...
Given information:
- Biased die is rolled 10 times.
- Probability of even number appearing 5 times is twice the probability of an even number appearing 4 times.
To find:
- Probability of an even number appearing twice when the die is rolled 8 times.
Solution:
Let's assume the probability of getting an even number on one roll of the die is 'p'.
Then, the probability of getting an odd number on one roll of the die is '1-p'.
Now, we need to find the probability of getting an even number twice when the die is rolled 8 times.
This can be calculated using the binomial probability formula:
P(X=k) = nCk * p^k * (1-p)^(n-k)
where,
- P(X=k) is the probability of getting k successes (even numbers appearing) in n trials (rolls of the die).
- nCk is the binomial coefficient, which represents the number of ways of choosing k successes from n trials.
- p^k is the probability of k successes (even numbers appearing) in one trial (roll of the die).
- (1-p)^(n-k) is the probability of (n-k) failures (odd numbers appearing) in n trials (rolls of the die).
Using the given information, we can set up two equations:
- P(X=5) = 2 * P(X=4) (probability of getting an even number 5 times is twice the probability of getting an even number 4 times)
- P(X=10) = 1 (total probability of rolling the die 10 times)
Using these equations, we can solve for 'p' and 'n':
- P(X=5) = 2 * P(X=4)
=> 10C5 * p^5 * (1-p)^5 = 2 * 10C4 * p^4 * (1-p)^6
=> 252 * p^5 * (1-p)^5 = 2 * 210 * p^4 * (1-p)^6
=> 21 * p * (1-p) = 2 * (1-p)^2
=> 21 * p = 2 * (1-p)
=> p = 2/23
- P(X=10) = 1
=> 8C0 * (2/23)^0 * (21/23)^8 + 8C1 * (2/23)^1 * (21/23)^7 + ... + 8C8 * (2/23)^8 * (21/23)^0 = 1
Now, we can use the binomial probability formula to find the probability of getting an even number twice when the die is rolled 8 times:
P(X=2) = 8C2 * (2/23)^2 * (21/23)^6
=> 28 * (4/529) * (410338673/16777216)
=> 0.0304
Therefore, the probability of getting an even number twice when the die is rolled 8 times is 0.0304, which is option (A).
- Biased die is rolled 10 times.
- Probability of even number appearing 5 times is twice the probability of an even number appearing 4 times.
To find:
- Probability of an even number appearing twice when the die is rolled 8 times.
Solution:
Let's assume the probability of getting an even number on one roll of the die is 'p'.
Then, the probability of getting an odd number on one roll of the die is '1-p'.
Now, we need to find the probability of getting an even number twice when the die is rolled 8 times.
This can be calculated using the binomial probability formula:
P(X=k) = nCk * p^k * (1-p)^(n-k)
where,
- P(X=k) is the probability of getting k successes (even numbers appearing) in n trials (rolls of the die).
- nCk is the binomial coefficient, which represents the number of ways of choosing k successes from n trials.
- p^k is the probability of k successes (even numbers appearing) in one trial (roll of the die).
- (1-p)^(n-k) is the probability of (n-k) failures (odd numbers appearing) in n trials (rolls of the die).
Using the given information, we can set up two equations:
- P(X=5) = 2 * P(X=4) (probability of getting an even number 5 times is twice the probability of getting an even number 4 times)
- P(X=10) = 1 (total probability of rolling the die 10 times)
Using these equations, we can solve for 'p' and 'n':
- P(X=5) = 2 * P(X=4)
=> 10C5 * p^5 * (1-p)^5 = 2 * 10C4 * p^4 * (1-p)^6
=> 252 * p^5 * (1-p)^5 = 2 * 210 * p^4 * (1-p)^6
=> 21 * p * (1-p) = 2 * (1-p)^2
=> 21 * p = 2 * (1-p)
=> p = 2/23
- P(X=10) = 1
=> 8C0 * (2/23)^0 * (21/23)^8 + 8C1 * (2/23)^1 * (21/23)^7 + ... + 8C8 * (2/23)^8 * (21/23)^0 = 1
Now, we can use the binomial probability formula to find the probability of getting an even number twice when the die is rolled 8 times:
P(X=2) = 8C2 * (2/23)^2 * (21/23)^6
=> 28 * (4/529) * (410338673/16777216)
=> 0.0304
Therefore, the probability of getting an even number twice when the die is rolled 8 times is 0.0304, which is option (A).
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