Which of the following three events is more likely to happen that a person get a) At least 1 six when 6 dice are rolled; b) At least 2 sixes when 12 dice are rolled, or c) At least 3sixes when 18 dice are rolled. Explain why.

prob(getting a six) = 1/6

prob(not getting a six) = 5/6

a) prob (at least 1 six in 6 tries)
---> we don't want all non-sixes
prob(no six in 6 tries) = C(6,0)(1/6)^0 (5/6)^6
= .3349
so prob of at least 1 six = 1 - .3349 = .6651

b) this time we exclude the case of no sixes + 1 six from 12
the excluded case
= C(12,0)(1/6)^0 (5/6)^11 + C(12,1)(1/6) (5/6)^10
= .1346 + .3230
prob(at least 2 sixes in 12 tries)
= 1 - .1346 - .3230 = .5424

c) use the same argument as I used above ....
then compare the 3 probabilities

To determine which of the three events is more likely to happen, we need to calculate the probabilities for each event.

Let's start with event a) - getting at least 1 six when 6 dice are rolled. To find this probability, we can calculate the complementary event, which is the probability of not getting any sixes and subtract it from 1.

The probability of not getting a six on a single die roll is 5/6 (since there are 5 possible outcomes that are not a six, out of 6 total possible outcomes). Since each die roll is independent, the probability of not getting a six on any of the 6 dice rolls is (5/6) raised to the power of 6. Therefore, the probability of getting at least 1 six when 6 dice are rolled is 1 - ((5/6)^6) ≈ 0.665.

Moving on to event b) - getting at least 2 sixes when 12 dice are rolled. Similarly, we need to calculate the complementary event of getting less than 2 sixes and subtract it from 1.

The probability of getting less than 2 sixes when rolling 12 dice can be calculated by adding the individual probabilities of getting 0 and 1 six. The probability of not getting a six on a single die roll is 5/6, as we calculated previously. Therefore, the probability of getting 0 sixes on 12 dice rolls is ((5/6)^12). The probability of getting 1 six is the number of ways to choose 1 die to be a six (12 choose 1) multiplied by the probability of getting a six on that chosen die (1/6), multiplied by the probability of not getting a six on the remaining 11 dice ((5/6)^11).

Putting it all together, the probability of getting less than 2 sixes when rolling 12 dice is ((5/6)^12) + (12 choose 1) * (1/6) * ((5/6)^11). Subtracting this probability from 1 gives us the probability of getting at least 2 sixes, which is approximately 0.368.

Finally, we have event c) - getting at least 3 sixes when 18 dice are rolled. Again, we calculate the complementary event of getting less than 3 sixes and subtract it from 1.

The probability of getting less than 3 sixes when rolling 18 dice can be calculated similarly to event b). We can calculate the probability of getting 0, 1, and 2 sixes, and then subtract this probability from 1 to get the probability of getting at least 3 sixes. The formula becomes (1 - ((5/6)^18) - (18 choose 1) * (1/6) * ((5/6)^17) - (18 choose 2) * (1/6^2) * ((5/6)^16)). The probability of getting at least 3 sixes is approximately 0.196.

Comparing the probabilities, we find that event a) has the highest probability at approximately 0.665, followed by event b) at approximately 0.368, and event c) has the lowest probability at approximately 0.196. Therefore, the most likely event to happen is getting at least 1 six when 6 dice are rolled.