Answer:
n = 60
Step-by-step explanation:
Total outcomes, while tossing a coin are 2, given as:
{Heads, Tails}
The probability of getting a heads up, while tossing the coin is:
P(Heads) = Favorable Outcome/Total Outcomes
P(Heads) = 1/2 = 0.5
Now, the total outcomes, while rolling a die are 6, given as:
{1, 2, 3, 4, 5, 6}
5 of these numbers are factors of 12, which are:
{1, 2, 3, 4, 6}
Thus, the probability of getting a factor of 12, while rolling a die is:
P(Factor of 12) = Favorable Outcome/Total Outcomes
P(Factor of 12) = 5/6 = 0.83
So, the probability of getting both heads and factor of 12 will be:
P(Heads and Factor of 12) = P(Heads ∪ Factor of 12)
P(Heads and Factor of 12) = P(Heads) * P(Factor of 12) = 0.5 * 0.83
P(Heads and Factor of 12) = 0.417 = 41.7%
So, for 144 trials, the number of trials in which we get heads and factor of 12, are given by:
n = P(Heads and Factor of 12) * 144
where,
n = no. of trials John would expect to roll a number on the die that is a factor of 12 and toss a coin that lands heads-up out of next 144 trials
n = 0.417 * 144
<u>n = 60</u>
