What's the answer for these probability problems?

Question

A coin is tossed twice and a die is rolled once. Find the probability that at least 1 head is tossed and a 5 does not show on the die.

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.

Answer 1

The coin is tossed twice. It has a 1/4 chance of being tails both times. The other 3/4 is the chance that it is not tails both times, which means heads showed up at least once.

There is a 5/6 chance that a 5 does not show on the die.

Multiply those together, and it's 3/4 * 5/6, which is 5/8.

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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.

Answer 2

we know that there are 6 sided of dice.
therefore 6 probabilities arise with the dice
as far the coin is considered,there are 2 probabilities
with these both, there are totally 2*6=12 probabilities.
but we want at least 1 head when 5 does not show on the die
therefore number of probabilities become head and 5 sides except 5=1*5=5 probabilities
therefore probability=5/12

second case
we want least 1 head and a number less than 4 shows on the die

number of probabilities=1*3=3

therefore the probability=3/12

Answer 3

3/4 * 5/6 = 15/24 = 5/8

3/4 * 1/2 = 3/8

Nice...Nissi