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

$\begin{gathered} \text{Probability of white = }\frac{90}{360}=\frac{1}{4} \\ \text{Probability of blue = }\frac{90}{360}=\frac{1}{4} \\ \text{Probability of red = }\frac{180}{360}=\frac{1}{2} \end{gathered}$

Part A

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

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

$\begin{gathered} \text{Probability(WB)=}\frac{1}{4}\times\frac{1}{4} \\ \text{Probability(WB)=}\frac{1}{16} \end{gathered}$

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

$\begin{gathered} P(WR)=\frac{1}{4}\times\frac{1}{2} \\ P(WR)=\frac{1}{8} \end{gathered}$

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)

$\begin{gathered} P(WW)=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16} \\ P(WB)=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16} \\ P(WR)=\frac{1}{4}\times\frac{1}{2}=\frac{1}{8} \\ P(BW)=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16} \\ P(BR)=\frac{1}{4}\times\frac{1}{2}=\frac{1}{8} \\ P(RW)=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8} \\ P(RB)=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8} \\ P(RR)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4} \end{gathered}$

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

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

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

$\begin{gathered} P(\text{not BB) = 1-P(BB)} \\ P(BB)=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16} \\ P(\text{not BB) = 1-}\frac{1}{16} \\ P(\text{not BB)=}\frac{15}{16} \end{gathered}$

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