Question

there are 7 Red 3 blue and 5 green marbles in a bag what is the probability that the first three chosen will not be red?

Answer 1

Determine the total number of marble in the bag.

$\begin{gathered} n(T)=7+3+5 \\ =15 \end{gathered}$

Determine the probability for first three marbles be red.

$\begin{gathered} P(R=3)=\frac{^7C_3^{}}{^{15}C_3} \\ =\frac{35}{455} \\ =\frac{7}{91} \end{gathered}$

The probability for first three marble is not to be red is equal to one minus the probability for the first three marble be red.

$\begin{gathered} P(R\ne3)=1-P(R=3) \\ =1-\frac{7}{91} \\ =\frac{91-7}{91} \\ =\frac{84}{91} \\ =\frac{12}{13} \end{gathered}$

So answer is 12/13.