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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:
- Rolling a biased die 10 times
- Probability of getting an even number 5 times is twice the probability of getting an even number 4 times
To find:
- Probability of getting an even number twice in 8 rolls
Solution:
Let P(5) be the probability of getting an even number 5 times in 10 rolls
Let P(4) be the probability of getting an even number 4 times in 10 rolls
Then, we can write:
P(5) = 2P(4) ......(1)
Now, we need to find the probability of getting an even number twice in 8 rolls.
Let's break this down into cases:
- Case 1: even number appears twice in first 2 rolls
- Case 2: even number appears once in first 2 rolls and once in next 6 rolls
- Case 3: even number appears once in last 2 rolls and once in first 6 rolls
Probability of even number appearing twice in first 2 rolls:
P(E) = Probability of getting an even number = 1/2 (assuming the die is biased towards even numbers)
P(NE) = Probability of not getting an even number = 1 - P(E) = 1/2
So, the probability of getting an even number twice in first 2 rolls = P(E) x P(E) = (1/2) x (1/2) = 1/4
Probability of even number appearing once in first 2 rolls and once in next 6 rolls:
P(E) = Probability of getting an even number = 1/2 (assuming the die is biased towards even numbers)
P(NE) = Probability of not getting an even number = 1 - P(E) = 1/2
So, the probability of getting an even number once in first 2 rolls and once in next 6 rolls = P(E) x P(NE) x P(E) x P(NE) x P(E) x P(NE) x P(E) x P(NE) = (1/2)^4 = 1/16
Probability of even number appearing once in last 2 rolls and once in first 6 rolls:
P(E) = Probability of getting an even number = 1/2 (assuming the die is biased towards even numbers)
P(NE) = Probability of not getting an even number = 1 - P(E) = 1/2
So, the probability of getting an even number once in last 2 rolls and once in first 6 rolls = P(E) x P(NE) x P(E) x P(NE) x P(E) x P(NE) x P(E) x P(NE) = (1/2)^4 = 1/16
Adding up the probabilities from all cases, we get:
P(getting an even number twice in 8 rolls) = 1/4 + 1/16 + 1/16 = 8/64 = 1/8
Now, we can use equation (1) to find P(4):
P(5) = 2P(4)
P(4) = P(5)/2 = 1/3
Finally, we can use the binomial distribution formula to find the probability of getting an even number twice in 8
- Rolling a biased die 10 times
- Probability of getting an even number 5 times is twice the probability of getting an even number 4 times
To find:
- Probability of getting an even number twice in 8 rolls
Solution:
Let P(5) be the probability of getting an even number 5 times in 10 rolls
Let P(4) be the probability of getting an even number 4 times in 10 rolls
Then, we can write:
P(5) = 2P(4) ......(1)
Now, we need to find the probability of getting an even number twice in 8 rolls.
Let's break this down into cases:
- Case 1: even number appears twice in first 2 rolls
- Case 2: even number appears once in first 2 rolls and once in next 6 rolls
- Case 3: even number appears once in last 2 rolls and once in first 6 rolls
Probability of even number appearing twice in first 2 rolls:
P(E) = Probability of getting an even number = 1/2 (assuming the die is biased towards even numbers)
P(NE) = Probability of not getting an even number = 1 - P(E) = 1/2
So, the probability of getting an even number twice in first 2 rolls = P(E) x P(E) = (1/2) x (1/2) = 1/4
Probability of even number appearing once in first 2 rolls and once in next 6 rolls:
P(E) = Probability of getting an even number = 1/2 (assuming the die is biased towards even numbers)
P(NE) = Probability of not getting an even number = 1 - P(E) = 1/2
So, the probability of getting an even number once in first 2 rolls and once in next 6 rolls = P(E) x P(NE) x P(E) x P(NE) x P(E) x P(NE) x P(E) x P(NE) = (1/2)^4 = 1/16
Probability of even number appearing once in last 2 rolls and once in first 6 rolls:
P(E) = Probability of getting an even number = 1/2 (assuming the die is biased towards even numbers)
P(NE) = Probability of not getting an even number = 1 - P(E) = 1/2
So, the probability of getting an even number once in last 2 rolls and once in first 6 rolls = P(E) x P(NE) x P(E) x P(NE) x P(E) x P(NE) x P(E) x P(NE) = (1/2)^4 = 1/16
Adding up the probabilities from all cases, we get:
P(getting an even number twice in 8 rolls) = 1/4 + 1/16 + 1/16 = 8/64 = 1/8
Now, we can use equation (1) to find P(4):
P(5) = 2P(4)
P(4) = P(5)/2 = 1/3
Finally, we can use the binomial distribution formula to find the probability of getting an even number twice in 8
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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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