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

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

=  $\frac{1}{6}$

Answer 2

The probability of getting sum 2 is $\frac{1}{6}$

The probability of getting sum 3 is $\frac{1}{3}$

The probability of getting sum 4 is $\frac{1}{3}$

The probability of getting sum 5 is $\frac{1}{6}$

Probability :

Two spinners shown in the question.

First spinner is divide into two equal parts.

Second spinner is divided into three equal parts.

The probability of getting sum 2 is,

             $P(sum=2)=\frac{1}{2} *\frac{1}{3}\\ \\P(sum=2)=\frac{1}{6}$

The probability of getting sum 3 is,

             $P(sum=3)=(\frac{1}{2}*\frac{1}{3}) +(\frac{1}{2}*\frac{1}{3})\\\\P(sum=3)=\frac{1}{3}$

The probability of getting sum 4 is,

             $P(sum=4)=(\frac{1}{2}*\frac{1}{3}) +(\frac{1}{2}*\frac{1}{3})\\\\P(sum=4)=\frac{1}{3}$

The probability of getting sum 5 is,

             $P(sum=5)=\frac{1}{2} *\frac{1}{3}\\ \\P(sum=5)=\frac{1}{6}$

Learn more about the Probability here:

https://brainly.com/question/24756209