Question

Juan spins the spinner shown below. He notes the area on which the arrow stops, and then spins it a second time. Find the probability that the outcome of the two spins is:a) red both timesb) white first, blue secondc) white first, red secondd) not blue both timesspinnerrwbAnswer

Answer 1

Given:

From the figure,

the sectoral angle of white = 90 degrees

the sectoral angle of blue = 90 degrees

the sectoral angle of red = 180 degrees

Total = 360 degrees

`\begin{gathered} \text{Probability of white = }(90)/(360)=(1)/(4) \\ \text{Probability of blue = }(90)/(360)=(1)/(4) \\ \text{Probability of red = }(180)/(360)=(1)/(2) \end{gathered}`

Part A

`\begin{gathered} \text{Probability of red both times } \\ P(R\text{ R)= }(1)/(2)*(1)/(2) \\ P(R\text{ R)=}(1)/(4) \end{gathered}`

Therefore, the probability that the outcome of the two spins is red both times is 1/4.

Part B

Probability of white first, blue second is

`\begin{gathered} \text{Probability(WB)=}(1)/(4)*(1)/(4) \\ \text{Probability(WB)=}(1)/(16) \end{gathered}`

Therefore, the probability that the outcome of the two spins is white first, blue second is 1/16.

Part C

Probability of white first, red second is

`\begin{gathered} P(WR)=(1)/(4)*(1)/(2) \\ P(WR)=(1)/(8) \end{gathered}`

Therefore, the probability that the outcome of the two spins is white first, red second is 1/8.

Part D

To get the probability of not blue both times, we have the following possibilities

P(WW or WB or WR or BW or BR or RW or RB or RR)

`\begin{gathered} P(WW)=(1)/(4)*(1)/(4)=(1)/(16) \\ P(WB)=(1)/(4)*(1)/(4)=(1)/(16) \\ P(WR)=(1)/(4)*(1)/(2)=(1)/(8) \\ P(BW)=(1)/(4)*(1)/(4)=(1)/(16) \\ P(BR)=(1)/(4)*(1)/(2)=(1)/(8) \\ P(RW)=(1)/(2)*(1)/(4)=(1)/(8) \\ P(RB)=(1)/(2)*(1)/(4)=(1)/(8) \\ P(RR)=(1)/(2)*(1)/(2)=(1)/(4) \end{gathered}`

Hence, the probability that it is not blue is the sum of all the individual probabilities above

`\begin{gathered} =(1)/(16)+(1)/(16)+(1)/(8)+(1)/(16)+(1)/(8)+(1)/(8)+(1)/(8)+(1)/(4) \\ =(1+1+2+1+2+2+2+4)/(16) \\ =(15)/(16) \\ P(\text{not blue both times)= }(15)/(16) \end{gathered}`

Alternatively, the probability of not blue both times can be given by;

`\begin{gathered} P(\text{not BB) = 1-P(BB)} \\ P(BB)=(1)/(4)*(1)/(4)=(1)/(16) \\ P(\text{not BB) = 1-}(1)/(16) \\ P(\text{not BB)=}(15)/(16) \end{gathered}`

Therefore, the probability that the outcome of the two spins is not blue both times is 15/16.