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Suppose a fair six-sided die is rolled once, if the value on the die is 1, 2 or 3 then die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
- a)10/21
- b)5/12
- c)2/3
- d)1/6
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Suppose a fair six-sided die is rolled once, if the value on the die i...
If first throw is 1, 2 or 3 then sample space is only 18 possible ordered pairs. Out of this only (1, 5 ), (1, 6 ), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5) and (3,6) i e. 9 out of 18 ordered pairs gives a Sum ≥ 6.
If first throw is 4, 5 or 6 then second throw is not made and therefore the only way Sum ≥ 6 is if the throw was 6. Which is one out of 3 possible. So the tree diagram becomes as follows:
From above diagram
If first throw is 4, 5 or 6 then second throw is not made and therefore the only way Sum ≥ 6 is if the throw was 6. Which is one out of 3 possible. So the tree diagram becomes as follows:
From above diagram
Most Upvoted Answer
Suppose a fair six-sided die is rolled once, if the value on the die i...
Solution:
Given, a fair six-sided die is rolled once and if the value on the die is 1, 2 or 3 then a die is rolled a second time.
Let's find the sample space of rolling the die:
Sample space, S = {1, 2, 3, 4, 5, 6}
Now, let's find the possible outcomes when the die is rolled twice and the sum of values that turn up is at least 6.
Possible outcomes:
- When the first roll is 1: (1, 5), (1, 6)
- When the first roll is 2: (2, 4), (2, 5), (2, 6)
- When the first roll is 3: (3, 3), (3, 4), (3, 5), (3, 6)
Therefore, the total number of possible outcomes is 2 + 3 + 4 = 9
Now, let's find the probability of getting at least 6 when the die is rolled:
P(sum is at least 6) = number of favorable outcomes / total number of outcomes
P(sum is at least 6) = 9 / 18 (since the die is rolled twice with a probability of 1/2 for the first roll)
P(sum is at least 6) = 1/2
Therefore, the correct option is (B) 5/12.
Given, a fair six-sided die is rolled once and if the value on the die is 1, 2 or 3 then a die is rolled a second time.
Let's find the sample space of rolling the die:
Sample space, S = {1, 2, 3, 4, 5, 6}
Now, let's find the possible outcomes when the die is rolled twice and the sum of values that turn up is at least 6.
Possible outcomes:
- When the first roll is 1: (1, 5), (1, 6)
- When the first roll is 2: (2, 4), (2, 5), (2, 6)
- When the first roll is 3: (3, 3), (3, 4), (3, 5), (3, 6)
Therefore, the total number of possible outcomes is 2 + 3 + 4 = 9
Now, let's find the probability of getting at least 6 when the die is rolled:
P(sum is at least 6) = number of favorable outcomes / total number of outcomes
P(sum is at least 6) = 9 / 18 (since the die is rolled twice with a probability of 1/2 for the first roll)
P(sum is at least 6) = 1/2
Therefore, the correct option is (B) 5/12.
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Suppose a fair six-sided die is rolled once, if the value on the die is 1, 2 or 3 then die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?a)10/21b)5/12c)2/3d)1/6Correct answer is option 'B'. Can you explain this answer?
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