A coin is tossed twice and a die is rolled once. Find the probability that at most 1 head is tossed and a number less than 4 shows.

At most one head is tossed: this means that it did not yield heads on both tosses. There is a 1/4 chance that it is heads on both tosses, and a 3/4 chance remains for at most one head being tossed.

There are three numbers less than 4, so there is a 3/6, or 1/2, chance of a number less than 4 being rolled.

3/4 * 1/2 = 3/8.

uh better go with cases,atmost 1 head says either head or no head,by probablity 1st coin total case when head(h) tail (t) are tt hh th ht so we need either tt or ht probabilty=3/4 now we need a number less than 4 that is 1 2 or 3 so probabilty = 3/6 hence final answer wen both r satisfied ie 3/6*3/4 = 3/8