If two dice are simultaneously rolled n total number of possible outcomes is 6*6 =36.
We can get 11 as sum in se cases {(5,6), (6,5)}. So, probability of getting sum 11 is 2/36 = 1/18.

If two coins are tossed simultaneously total number of possible outcomes are 4.
{HH, TT, HT, TH}

Event ?A but not B? contains elements which are present in A but not in B.
It is represented by A-B or A∩B?.

There are six possible outcomes on dice i.e. 1 to 6.
Even numbers on dice are 2,4,6 i.e. three outcomes in favour of event.
So, probability of getting an even number on dice is 3/6 = 1/2.

A = {(1,1), (1,2), (2,1)}
B= {(1,1), (1,2), (1,3), (2,1), (2,2), (3,1)}
?A and B? contains elements present in both A and B.
So, A and B = {(1,1), (1,2), (2,1)}.

Since probability is ratio of number of outcomes in favour to total number of possible outcomes. And number of outcomes in favour are always less than or equal to total number of possible outcomes. So, probability is always less than or equal to 1.

If two coins are tossed simultaneously n S = {HH, HT, TH, TT}.
?Exactly one head appears? includes HT, TH. It has more than one sample point so it is compound event not simple event.

If two dice are simultaneously rolled n total number of possible outcomes is 6*6 =36.
We can get 4 as sum in se cases {(1,3), (2,2), (3,1)}. So, probability of getting sum 4 is 3/36 = 1/12.

Each bulb can have two cases i.e. eir defective or non-defective. Then total number of outcomes will be 2⁴ = 16.
Probability that all bulbs are defective is 1/16.
So, probability that at least one bulb is non-defective is 1-1/16 = 15/16.

If two coins are tossed simultaneously total number of possible outcomes are 4.
{HH, TT, HT, TH} out of which {HT, TH, TT} favour event.
So, probability of getting at most one head = 3/4.