Question

A box contains 6 red and 4 white marbles. A marble is drawn and replaced three times from the box. The probability that one white marbles is drawn.
A. $\dfrac{53}{125}$
B. $\dfrac{54}{125}$
C. $\dfrac{56}{125}$
D. $\dfrac{52}{125}$

Answer 1

To obtain the probability of one white marble is drawn we will take all the condition in which the three balls are drawn from the box. Firstly we will find the total number of balls in the box then we will take cases in which the ball can be drawn and finally by using Combination we will get the desired answer.

Complete step by step answer:
It is given to us that the box has following balls:
Red- $R=6$
White- $W=4$
So the total numbers of balls in the box is:
Total number of balls $=6+4$
Total number of balls $=10$
Now the ways in which the three balls can be drawn such that we get one white ball are as follows:
$RRW,RWR,WRR$
So the probability that one white ball is drawn will be:
$P$(One white Ball) $=\left( \dfrac{{}^{6}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)\times \left( \dfrac{{}^{6}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)\times \left( \dfrac{{}^{4}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)+\left( \dfrac{{}^{6}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)\times \left( \dfrac{{}^{4}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)\times \left( \dfrac{{}^{6}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)+\left( \dfrac{{}^{4}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)\times \left( \dfrac{{}^{6}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)\times \left( \dfrac{{}^{6}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)$
$P$(One white Ball) $=3\left( \dfrac{{}^{6}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)\times \left( \dfrac{{}^{6}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)\times \left( \dfrac{{}^{4}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)$
Using combination formula that states:
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Using the above formula we get,
$P$(One white Ball) $=3\left( \dfrac{\dfrac{6!}{1!\left( 6-1 \right)!}}{\dfrac{10!}{1!\left( 10-1 \right)!}} \right)\times \left( \dfrac{\dfrac{6!}{1!\left( 6-1 \right)!}}{\dfrac{10!}{1!\left( 10-1 \right)!}} \right)\times \left( \dfrac{\dfrac{4!}{1!\left( 4-1 \right)!}}{\dfrac{10!}{1!\left( 10-1 \right)!}} \right)$
$P$(One white Ball) $=3\left( \dfrac{6}{10} \right)\times \left( \dfrac{6}{10} \right)\times \left( \dfrac{4}{10} \right)$
$P$(One white Ball) $=3\times \dfrac{3}{5}\times \dfrac{3}{5}\times \dfrac{2}{5}$
$P$(One white Ball) $=\dfrac{54}{125}$
So the probability of drawing one white ball is $\dfrac{54}{125}$

So, the correct answer is “Option B”.

Note: Probability is used to find out how likely an event is to occur or how likely it is for a proposition to be true. The higher the probability of an event the more chances are there that the event will occur. Probability of an event lies between 0 and 1 it can’t be greater than 1 and neither can it be negative.