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A pair of fair dice are rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is
- a)0.45
- b)0.4
- c)0.5
- d)0.7
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A pair of fair dice are rolled together till a sum of either 5 or 7 is...
We do not have to consider any sum other than 5 or 7 occurring.
A sum of 5 can be obtained by any of [4 + 1, 3 + 2, 2 + 3, 1 + 4]
Similarly a sum of 7 can be obtained by any of [6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5, 1 + 6]
For 6: n(E) = 4, n(S) = 6 + 4 P = 0.4
For 7: n(E) = 6 n(S) = 6 + 4 P = 0.6
A sum of 5 can be obtained by any of [4 + 1, 3 + 2, 2 + 3, 1 + 4]
Similarly a sum of 7 can be obtained by any of [6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5, 1 + 6]
For 6: n(E) = 4, n(S) = 6 + 4 P = 0.4
For 7: n(E) = 6 n(S) = 6 + 4 P = 0.6
Most Upvoted Answer
A pair of fair dice are rolled together till a sum of either 5 or 7 is...
Solution:
When two dice are rolled together, there are a total of 36 possible outcomes. The sum of the numbers appearing on the dice will be from 2 to 12.
Finding the Probability of obtaining sum of 5 or 7:
We need to find the probability of obtaining a sum of either 5 or 7 on rolling a pair of dice.
- Sum of 5: There are four ways to get a sum of 5 (1+4, 2+3, 3+2, 4+1)
- Sum of 7: There are six ways to get a sum of 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1)
Therefore, the probability of getting a sum of either 5 or 7 is 4/36 + 6/36 = 10/36 = 5/18.
Finding the Probability of obtaining 5 before 7:
We need to find the probability of obtaining a sum of 5 before a sum of 7.
Let P be the required probability.
- If the first roll is a sum of 5, then the game ends and we win. The probability of this happening is 4/36.
- If the first roll is a sum of 7, then the game ends and we lose. The probability of this happening is 6/36.
- If the first roll is a sum other than 5 or 7, then we ignore it and continue rolling until we get a sum of either 5 or 7. At this point, we either win (if the sum is 5) or lose (if the sum is 7).
Let Q be the probability of winning from this point onwards (i.e. getting a sum of 5 before 7).
- If the sum of the first roll is 6, then the only way to win is to get a sum of 5 on the second roll. The probability of this happening is 4/36.
- If the sum of the first roll is 8, then the only way to win is to get a sum of 5 on the second roll. The probability of this happening is 4/36.
- If the sum of the first roll is 9, then the only way to win is to get a sum of 5 on the second roll. The probability of this happening is 4/36.
- If the sum of the first roll is 10, then there are two ways to win: either by getting a sum of 5 on the second roll (probability of 3/36) or by getting a sum of 7 on the second roll, followed by a sum of 5 on the third roll (probability of 2/36). Therefore, the probability of winning from this point onwards is (3/36)+(2/36) = 5/36.
- If the sum of the first roll is 11, then the only way to win is to get a sum of 5 on the second roll. The probability of this happening is 4/36.
- If the sum of the first roll is 12, then there are two ways to win: either by getting a sum of 7 on the second roll, followed by a sum of 5 on the
When two dice are rolled together, there are a total of 36 possible outcomes. The sum of the numbers appearing on the dice will be from 2 to 12.
Finding the Probability of obtaining sum of 5 or 7:
We need to find the probability of obtaining a sum of either 5 or 7 on rolling a pair of dice.
- Sum of 5: There are four ways to get a sum of 5 (1+4, 2+3, 3+2, 4+1)
- Sum of 7: There are six ways to get a sum of 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1)
Therefore, the probability of getting a sum of either 5 or 7 is 4/36 + 6/36 = 10/36 = 5/18.
Finding the Probability of obtaining 5 before 7:
We need to find the probability of obtaining a sum of 5 before a sum of 7.
Let P be the required probability.
- If the first roll is a sum of 5, then the game ends and we win. The probability of this happening is 4/36.
- If the first roll is a sum of 7, then the game ends and we lose. The probability of this happening is 6/36.
- If the first roll is a sum other than 5 or 7, then we ignore it and continue rolling until we get a sum of either 5 or 7. At this point, we either win (if the sum is 5) or lose (if the sum is 7).
Let Q be the probability of winning from this point onwards (i.e. getting a sum of 5 before 7).
- If the sum of the first roll is 6, then the only way to win is to get a sum of 5 on the second roll. The probability of this happening is 4/36.
- If the sum of the first roll is 8, then the only way to win is to get a sum of 5 on the second roll. The probability of this happening is 4/36.
- If the sum of the first roll is 9, then the only way to win is to get a sum of 5 on the second roll. The probability of this happening is 4/36.
- If the sum of the first roll is 10, then there are two ways to win: either by getting a sum of 5 on the second roll (probability of 3/36) or by getting a sum of 7 on the second roll, followed by a sum of 5 on the third roll (probability of 2/36). Therefore, the probability of winning from this point onwards is (3/36)+(2/36) = 5/36.
- If the sum of the first roll is 11, then the only way to win is to get a sum of 5 on the second roll. The probability of this happening is 4/36.
- If the sum of the first roll is 12, then there are two ways to win: either by getting a sum of 7 on the second roll, followed by a sum of 5 on the
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A pair of fair dice are rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 isa)0.45b)0.4c)0.5d)0.7Correct answer is option 'B'. Can you explain this answer?
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