Question

A box contains $5$ red marbles, $8$ white marbles and $4$ green marbles. On marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red (ii) white (iii) not green

Answer 1

Total no. of marbles in the box is $17$ $(5$ red, $4$ green and $8$ white$)$

Solution(i):

No. of red marbles in box is $5$

Therefore, ${}^5{C}_1$( Selecting $1$ out of $5$ items) times out of ${}^{17}{C}_1$( Selecting $1$ out of $17$ items) a red marbles is picked.

Let $E$ be the event of drawing a red marble from the box

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^5{C}_1}{{}^{17}{C}_1}$$=\frac{5}{17}$

Solution(ii):

No. of white marbles in box is $8$

Therefore, ${}^8{C}_1$( Selecting $1$ out of $8$ items) times out of ${}^{17}{C}_1$( Selecting $1$ out of $17$ items) a white marbles is picked.

Let $E$ be the event of drawing a white marble from the box

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^8{C}_1}{{}^{17}{C}_1}$$=\frac{8}{17}$

Solution(iii):

No. of non-green marbles in box is $13$$(5$ red and $8$ white$)$

Therefore, ${}^{13}{C}_1$( Selecting $1$ out of $13$ items) times out of ${}^{17}{C}_1$( Selecting $1$ out of $17$ items) a non-green marbles is picked.

Let $E$ be the event of drawing a non-green marble from the box

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{13}{C}_1}{{}^{17}{C}_1}$$=\frac{13}{17}$