Answer:
3/16
Explanation:
• The total number of sections in the spinner, n(S)=8
,
• The number of gray sections in the spinner, n(G)= 6
,
• The number of blue sections in the spinner, n(B)= 2
The probability that the first spinner lands on gray(G) and the second lands on blue(B):
[tex]\begin{gathered} P(GB)=P(G)\times P(B) \\ =\frac{n(G)}{n(S)}\times\frac{n(B)}{n(S)} \\ =\frac{6}{8}\times\frac{2}{8} \\ =\frac{12}{64} \\ =\frac{3}{16} \end{gathered}[/tex]
The required probability is 3/16.
