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In a simultaneous throw of a pair of dice, the probability of getting a sum more than 7 will be
- a)5/12
- b)13/12
- c)7/12
- d)9/12
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
In a simultaneous throw of a pair of dice, the probability of getting ...
Here n(S) = (6 x 6) = 36
Let E = event of getting a total more than 7
= {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}
Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.
Most Upvoted Answer
In a simultaneous throw of a pair of dice, the probability of getting ...
There are 36 total number of outcome and 15 required outcome
=15/36
=5/12
hence the answer is correct
if u like my answer please upvote and follow me
=15/36
=5/12
hence the answer is correct
if u like my answer please upvote and follow me
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Community Answer
In a simultaneous throw of a pair of dice, the probability of getting ...
To find the probability of getting a sum more than 7 in a simultaneous throw of a pair of dice, we need to determine the number of favorable outcomes and the total number of possible outcomes.
Let's first identify the favorable outcomes:
When throwing two dice, there are 6 possible outcomes for each dice, resulting in a total of 6 * 6 = 36 possible outcomes.
For the sum to be more than 7, we need to consider the following favorable outcomes:
- When the first dice shows 1, the second dice can show any number from 6 to 1, resulting in 6 favorable outcomes.
- When the first dice shows 2, the second dice can show any number from 6 to 2, resulting in 5 favorable outcomes.
- When the first dice shows 3, the second dice can show any number from 6 to 3, resulting in 4 favorable outcomes.
- When the first dice shows 4, the second dice can show any number from 6 to 4, resulting in 3 favorable outcomes.
- When the first dice shows 5, the second dice can show any number from 6 to 5, resulting in 2 favorable outcomes.
- When the first dice shows 6, the second dice can only show 6, resulting in 1 favorable outcome.
Therefore, the total number of favorable outcomes is 6 + 5 + 4 + 3 + 2 + 1 = 21.
Now, let's calculate the total number of possible outcomes, which we have already determined as 36.
To find the probability, we divide the number of favorable outcomes by the total number of possible outcomes:
Probability = Favorable outcomes / Total outcomes
Probability = 21 / 36
Probability = 7 / 12
Therefore, the correct answer is option 'A' - 5/12.
Let's first identify the favorable outcomes:
When throwing two dice, there are 6 possible outcomes for each dice, resulting in a total of 6 * 6 = 36 possible outcomes.
For the sum to be more than 7, we need to consider the following favorable outcomes:
- When the first dice shows 1, the second dice can show any number from 6 to 1, resulting in 6 favorable outcomes.
- When the first dice shows 2, the second dice can show any number from 6 to 2, resulting in 5 favorable outcomes.
- When the first dice shows 3, the second dice can show any number from 6 to 3, resulting in 4 favorable outcomes.
- When the first dice shows 4, the second dice can show any number from 6 to 4, resulting in 3 favorable outcomes.
- When the first dice shows 5, the second dice can show any number from 6 to 5, resulting in 2 favorable outcomes.
- When the first dice shows 6, the second dice can only show 6, resulting in 1 favorable outcome.
Therefore, the total number of favorable outcomes is 6 + 5 + 4 + 3 + 2 + 1 = 21.
Now, let's calculate the total number of possible outcomes, which we have already determined as 36.
To find the probability, we divide the number of favorable outcomes by the total number of possible outcomes:
Probability = Favorable outcomes / Total outcomes
Probability = 21 / 36
Probability = 7 / 12
Therefore, the correct answer is option 'A' - 5/12.
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