Hey guys! I'm having trouble with this question. Thank you!
Two dice with their faces numbered 1 to 6 are tossed.
i. By listing the possible outcomes find the probability of
a) one die showing a 6 with other showing at least 4
b) one die showing a 6 with the other showing 3 or less
ii) Deduce from the results in part (i) the probability that neither die shows a 6.
The question asks to list the possibilities, hence you should. However I'll structure my answers as though they weren't listed
Let the rows correspond to the first dice, and the second correspond to the second dice.
The total possible outcomes is always 36
a) Suppose the first dice shows a 6. Then the second die shows either 4, 5 or 6. There are 3 possible choices (going down column 6)
Suppose the second dice shows a 6. Then the first die shows either 4, 5 or 6. There are 3 possible choices (going across row 6)
(6,6) is counted twice, so we deduct it once.
Therefore there are 5 favourable outcomes, so the answer is 5/36
b) Suppose again the first dice shows a 6. Then the second shows either 1, 2 or 3 instead. There are 3 possible choices
And similarly for the second dice
However here, nothing gets counted twice.
Therefore there are 6 favourable outcomes, so the answer is 6/36 = 1/6
For part ii), note that the sum of the above probabilities is 11/36.
Notice that in the above parts, we've catered for
all the cases where 6 must appear at least once.
Hence the probability 6 doesn't appear at all is just the complement. This has a probability of 25/36
This can also be deduced from reading off the table, as we are basically asking for the probability that something not in row 6 and/or column 6 is chosen.