Question

An experiment consists of tossing a die and then flipping a coin once if the number on the die is even. If the number on the die is odd, the coin is flipped twice. Using the notation 4H, for example, to denote the outcome that the die comes up 4 and then the coin comes up heads, and 3HT to denote the outcome that the die comes up 3 followed by a head and then a tail on the coin, construct the sample space S and then find the probability of getting an even number on the die followed by one head. 3/18 6/18 3/12 6/12

Answer 1

Answer:

$\Omega = \{ 1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T,\\5HH , 5HT, 5TH, 5TT, 6H, 6T \}$

The probability is 3/12. The third option is correct.

Step-by-step explanation:

The sample space is

$\Omega = \{ 1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T,\\5HH , 5HT, 5TH, 5TT, 6H, 6T \}$

Note that this sample space is not equally probable.

The probability of getting a given number followed is the probability of getting an even number from the 6 numbers (3/6) multiplied by the probability of getting a head after getting that even number, that is 1/2, because is equally probable to get heads or tails from one single coin toss (note that we are assuming that the dice was even, thats why there is a single coin toss).

Therefore, the probability of getting an even number and a head is

P( D in {2,4,6} , H = 1) = P(D in {2,4,6}) * P(H=1 | D in {2,4,6}) = 3/6 * 1/2 = 3/12.