Pregunta
Two fair dice are rolled 5 times and the sum of the numbers that come up is recorded. Find the probability of these events.
a. The sum is 7 on each of the five rolls.
b. The sum is 7 exactly four times in the five rolls.
a. The sum is 7 on each of the five rolls.
b. The sum is 7 exactly four times in the five rolls.
Ask by Bush Thornton. in the United States
Mar 22,2025
Solución de inteligencia artificial de Upstudy
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a. The sum is 7 on each of the five rolls
- Probability:
b. The sum is 7 exactly four times in the five rolls
- Probability:
Solución
a. The sum is 7 on each of the five rolls
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For two fair dice, the possible outcomes are
. -
The favorable outcomes for a sum of 7 are:
, , , , , and which gives outcomes. -
The probability of getting a sum of 7 in one roll is
-
Since the rolls are independent, the probability of getting a sum of 7 on each of the five rolls is
b. The sum is 7 exactly four times in the five rolls
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The probability of getting a sum of 7 in a single roll isand the probability of not getting a sum of 7 is
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We need exactly four successes (sum equals 7) out of five rolls. The number of ways to choose which four out of five rolls yield sum 7 is given by the combination
-
Hence, the probability is
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Extra Insights
When two fair dice are rolled, the possible sums range from 2 to 12, with some sums being more common than others. The sum of 7 is one of the most likely outcomes for two dice, as it can be achieved in 6 different combinations (1+6, 2+5, 3+4, 4+3, 5+2, and 6+1) out of a total of 36 possible combinations when rolling the dice. This gives a probability of rolling a sum of 7 as 6/36, which simplifies to 1/6.
To find the probability of rolling a sum of 7 on each of the five rolls, we use the formula for independent events. The probability of getting a sum of 7 on one roll is 1/6, so for five rolls, we compute (1/6)⁵. This yields a very small probability of (1/7776), which is equivalent to about 0.0129%.
Now, for getting a sum of 7 exactly four times in five rolls, we can apply the binomial probability formula. Here, we identify that there are 5 rolls, the number of successes (sum of 7) we want is 4, and the failure (not getting 7) in a single roll occurs with a probability of 5/6. Calculating it gives us:
This gives us a probability of 25/7776, around 0.00322, illustrating how getting close to perfection in probability must deal with that elusive, sneaky outcome!

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