Suppose that we roll three fair six-sided dice. (a) What is the probability that

Question

Suppose that we roll three fair six-sided dice.

(a) What is the probability that at least two of the dice will show a five or six?

(b) Suppose that we sum the numbers shown on the rolled dice. What is the probability that the sum will be a 15 or greater?

(c) How many times would we have to roll to have a better than 50% chance of getting at least 15? Remark: each time we roll three dice.

*Please post detailed solution, I need help gaining this concept*

(d) Finally, suppose that we consider only the two dice with the highest numbers. What is the probability that the sum of these two is 12?

Explanation / Answer

A fair six faced die has six faces and when throw the die all the outcomes are equally likely.

Three die thrown, the total number of possible outcomes are 6*6*6 =216

The favorable events to the outcome at least two of the dice will show a five or six is 56.

Those are (6,6,1) to (6,6,6) -(6) ; (5,5,,1) to (5,5,6) - (6); (6,5,1) to (6,5,6) - (6) ; (5,6,1) to (5,6,6) -(6); (6,1,6) to (6,4,6) -(4);

(6,1,5) to (6,4,5)-(4); (5,1,6) to (5,4,6) -(4); (5,1,5) to (5,4,5) -(4); (1,6,6) to (4,6,6) -(4); (1,6,5) to (4,6,5) -(4); (1,5,6) to (4,5,6)-(4); (1,5,5) to (4,5,5)-(4)

(a) The probability that at least two of the dice will show a five or six =56/216=0.2592

The sum of the rolled number varies from 3 to 18; Sum 15 or greater means

sum is 18, favorable no.of events is 1 (i.e 6,6,6); sum is 17, favorable no.of events are 3 (6,6,5),(6,5,6),(5,6,6)

sume is 16, favorbale no.of events are 6; sum is 15, favorable no.of events are 10

total no.of favorbale events are =19

(b) Suppose that we sum the numbers shown on the rolled dice then the probability that the sum will be a 15 or greater is

=20/216 = 0.092

(c) How many times would we have to roll to have a better than 50% chance of getting at least 15? Remark: each time we roll three dice

if you roll 1000 times 92 times would be at least 15. 50% of times should be at least 15 is 92/2 = 46

d) Finally, suppose that we consider only the two dice with the highest numbers. What is the probability that the sum of these two is 12?

If the two numbers with the highest numbers i.e 6 and 6 then their sum is 12. The minimum number is 1.

So this is an impossible event.

The probability that the sum of these two is 12 is '0'

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