Question

Rafael spins the pointers of the two spinners shown at the right. Find the probability of each possible sum

Answer 1

Step-by-step explanation:

• P(sum)= 2

On the left:  possible outcome has the value of 1 (1) = $\frac{1}{2}$

On the right: 1 possible outcome has the value of 1 (1)= $\frac{1}{3}$

=> the probability of possible P(sum)= 2

= $\frac{1}{2}$* $\frac{1}{3}$

=  $\frac{1}{6}$

• P(sum)= 3

On the left:  possible outcome has the value of 1  = $\frac{1}{2}$

On the right: 1 possible outcome has the value of 2 = $\frac{1}{3}$

On the left:  possible outcome has the value of 2  = $\frac{1}{2}$

On the right: 1 possible outcome has the value of 1 = $\frac{1}{3}$

=> the probability of possible P(sum)= 3

=  $\frac{1}{2}$*$\frac{1}{3}$  + $\frac{1}{2}$*$\frac{1}{3}$

= $\frac{1}{3}$

• P(sum)= 4

On the left:  possible outcome has the value of 1  = $\frac{1}{2}$

On the right: 1 possible outcome has the value of 3 = $\frac{1}{3}$

On the left:  possible outcome has the value of 2  = $\frac{1}{2}$

On the right: 1 possible outcome has the value of 2= $\frac{1}{3}$

=> the probability of possible P(sum)= 4

= $\frac{1}{2}$*$\frac{1}{3}$  + $\frac{1}{2}$*$\frac{1}{3}$

= $\frac{1}{3}$

• P(sum)= 5

On the left:  possible outcome has the value of 2  = $\frac{1}{2}$

On the right: 1 possible outcome has the value of 3 = $\frac{1}{3}$

=> the probability of possible P(sum)= 5

= $\frac{1}{2}$* $\frac{1}{3}$

=  $\frac{1}{6}$