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Answer:
11.57%
Step-by-step explanation:
The probability of rolling two 3s is the product of the probability of rolling two 3s and two not-3s, and the number of ways that combination of rolls can appear.
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<h3>combinations of rolls</h3>Of the four die rolls, we want two of them to be 3s. The 3s can appear anywhere in the sequence. For example, we could have 33xx, or xx33, or any of a number of other combinations (where x is a "not 3"). The total number of ways two 3s can appear an a sequence of 4 rolls is ...
C(4, 2) = 4!/(2!(4 -2)!) = 4·3/(2·1) = 6
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<h3>probability of two 3s and two not-3s</h3>The probability of rolling a 3 on a fair 6-sided die with one face labeled 3 is presumed to be 1/6. Then the probability of rolling something else is ...
1 -1/6 = 5/6
So, two 3s and two not-3s in a given sequence have a probability of ...
(1/6)(1/6)(5/6)(5/6) = 25/1296
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<h3>two 3s in 4 rolls</h3>The probability that four rolls of the die will show exactly two 3s is the product of the probability it can happen and the number of ways it can happen:
(25/1296) × 6 = 25/216 ≈ 11.57%
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The attachment shows the probability function of a calculator gives the same result.
I'm not sure how to show work on this problem but hopefully this helps.
