Question

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1. If each spinner is spun once, what is the probability that both spinners show the same color?
2. If each spinner is spun once, what is the probability of getting a red-blue combination?

Answer 1

`\sf 1. \quad (3)/(8)`

`\sf 2. \quad (7)/(24)`

Explanation:

Spinner 1

Spinner 1 is divided into 6 congruent sections.

There are 3 red sections, 2 blue sections and 1 yellow section.

Therefore, the probability of spinning each of the three colors is:

`\bullet \quad \sf P(R_1)=(3)/(6)=(1)/(2)`

`\bullet \quad \sf P(B_1)=(2)/(6)=(1)/(3)`

`\bullet \quad \sf P(Y_1)=(1)/(6)`

Spinner 2

Spinner 2 is divided into 3 sections of differing sizes.

The red section is half of the spinner. The blue and yellow sections are quarters of the spinner.

Therefore, the probability of spinning each of the three colors is:

`\bullet \quad \sf P(R_2)=(1)/(2)`

`\bullet \quad \sf P(B_2)=(1)/(4)`

`\bullet \quad \sf P(Y_2)=(1)/(4)`

Question 1

If each spinner is spun once, then:

The probability that both spinners both show red is:

`\sf P(R_1)\;and\;P(R_2)=(1)/(2) * (1)/(2)=(1)/(4)`

The probability that both spinners both show blue is:

`\sf P(B_1)\;and\;P(B_2)=(1)/(3) * (1)/(4)=(1)/(12)`

The probability that both spinners both show yellow is:

`\sf P(Y_1)\;and\;P(Y_2)=(1)/(6) * (1)/(4)=(1)/(24)`

Therefore, the probability that both spinners show the same colour is:

`\sf P(R_1\;\&\;R_2)\;or\;P(B_1\;\&\;B_2)\;or\;P(Y_1\;\&\;Y_2)=(1)/(4)+(1)/(12)+(1)/(24)=(9)/(24)=(3)/(8)`

Question 2

If each spinner is spun once, the probability of getting a red from spinner 1 and a blue from spinner 2 is:

`\sf P(R_1)\;and\;P(B_2)=(1)/(2) * (1)/(4)=(1)/(8)`

The probability of getting a blue from spinner 1 and a red from spinner 2 is:

`\sf P(B_1)\;and\;P(R_2)=(1)/(3) * (1)/(2)=(1)/(6)`

Therefore, the probability of getting a red-blue combination is:

`\sf P(R_1\;\&\;B_2)\;or\;P(B_1\;\&\;R_2)=(1)/(8)+(1)/(6)=(7)/(24)`