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A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If E is the event that a number greater than or equal to 4 occurs on a single toss of the die, then P (E) is
- a)1/2
- b)2/3
- c)4/9
- d)1/3
Correct answer is option 'C'. Can you explain this answer?
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A die is loaded in such a way that each odd number is twice as likely ...
Let probability of occurence of even number be k, then probability of occurence of odd number will be 2 k.
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A die is loaded in such a way that each odd number is twice as likely ...
Let probability of occurence of even number be k, then probability of occurence of odd number will be 2 k.
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A die is loaded in such a way that each odd number is twice as likely ...
Explanation:
To solve this problem, we need to determine the probability of rolling a number greater than or equal to 4 on a single toss of the loaded die, given that each odd number is twice as likely to occur as each even number.
Let's start by determining the probabilities of rolling each number on a fair, unbiased die:
- Probability of rolling a 1: 1/6
- Probability of rolling a 2: 1/6
- Probability of rolling a 3: 1/6
- Probability of rolling a 4: 1/6
- Probability of rolling a 5: 1/6
- Probability of rolling a 6: 1/6
Next, we need to consider the fact that each odd number is twice as likely to occur as each even number. This means that the probabilities of rolling odd numbers are doubled, while the probabilities of rolling even numbers remain the same.
- Probability of rolling an odd number: 2 * (1/6) = 2/6 = 1/3
- Probability of rolling an even number: 1/6
Now, we can determine the probability of rolling a number greater than or equal to 4. This includes rolling a 4, 5, or 6.
- Probability of rolling a 4: 1/6
- Probability of rolling a 5: 1/6
- Probability of rolling a 6: 1/6
Adding up these probabilities, we get:
P(E) = (1/6) + (1/6) + (1/6) = 3/6 = 1/2
However, this is the probability for a fair, unbiased die. Since we are dealing with a loaded die, where each odd number is twice as likely to occur as each even number, we need to adjust the probabilities accordingly.
Since the probability of rolling an odd number is 1/3 and the probability of rolling an even number is 1/6, the adjusted probability of rolling a number greater than or equal to 4 becomes:
P(E) = (1/3) + (1/3) + (1/3) = 3/3 = 1
Therefore, the correct answer is option 'C': 4/9.
To solve this problem, we need to determine the probability of rolling a number greater than or equal to 4 on a single toss of the loaded die, given that each odd number is twice as likely to occur as each even number.
Let's start by determining the probabilities of rolling each number on a fair, unbiased die:
- Probability of rolling a 1: 1/6
- Probability of rolling a 2: 1/6
- Probability of rolling a 3: 1/6
- Probability of rolling a 4: 1/6
- Probability of rolling a 5: 1/6
- Probability of rolling a 6: 1/6
Next, we need to consider the fact that each odd number is twice as likely to occur as each even number. This means that the probabilities of rolling odd numbers are doubled, while the probabilities of rolling even numbers remain the same.
- Probability of rolling an odd number: 2 * (1/6) = 2/6 = 1/3
- Probability of rolling an even number: 1/6
Now, we can determine the probability of rolling a number greater than or equal to 4. This includes rolling a 4, 5, or 6.
- Probability of rolling a 4: 1/6
- Probability of rolling a 5: 1/6
- Probability of rolling a 6: 1/6
Adding up these probabilities, we get:
P(E) = (1/6) + (1/6) + (1/6) = 3/6 = 1/2
However, this is the probability for a fair, unbiased die. Since we are dealing with a loaded die, where each odd number is twice as likely to occur as each even number, we need to adjust the probabilities accordingly.
Since the probability of rolling an odd number is 1/3 and the probability of rolling an even number is 1/6, the adjusted probability of rolling a number greater than or equal to 4 becomes:
P(E) = (1/3) + (1/3) + (1/3) = 3/3 = 1
Therefore, the correct answer is option 'C': 4/9.
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