Question

Fred flips a coin, spins the spinner, and rolls a standard number cube. Find the probability that the coin will show tails, the spinner will land on yellow or green, and the cube will show a three.

Answer 1

Answer:

1/12

Step-by-step explanation:

Let us begin by identifying all sample spaces

Sample space for coin =2

Sample space for spinner = 2

Sample space for die= 6

Total sample space S= 10

1. Probability that the coin will show tails, is = 1/2

2. Probability that the spinner will land on yellow or green, is

=1/2+1/2= 1

3. Probability that the cube will

show a three.= 1/6

Hence for this event that Fred carried out the probability is

1/2*1*1/6= 1/12

Answer 2

Answer:

The probability that the coin will show tails, the spinner will land on yellow or green (out of yellow, green, blue), and the cube will show a three is $\frac{4}{3}$  

Step-by-step explanation:

First we find probability one by one,

1) Flips a coin - Head, Tail

Total number of outcome = 2

Favorable outcome (getting a tail) = 1

Probability that the coin will show tails is $P(T)=\frac{1}{2}$

2) Spins the spinner - yellow, green, blue

Total number of outcome = 3

Favorable outcome (land on yellow or green) = 2

Probability that the spinner will land on yellow or green $P(S)=\frac{2}{3}$

3) Rolls a standard number cube - 1,2,3,4,5,6

Total number of outcome = 6

Favorable outcome (show a three) = 1

Probability that the cube will show a three $P(C)=\frac{1}{6}$

The probability that the coin will show tails, the spinner will land on yellow or green (out of yellow, green, blue), and the cube will show a three is

$P=P(T)+P(S)+P(C)\\\\P=\frac{1}{2}+ \frac{2}{3}+\frac{1}{6}\\\\P=\frac{3+4+1}{6}\\P=\frac{8}{6}\\P=\frac{4}{3}$

Therefore, The probability that the coin will show tails, the spinner will land on yellow or green (out of yellow, green, blue), and the cube will show a three is $\frac{4}{3}$