Question

A bowl has 6 red marbles and 3 green marbles. The probability that a blind folded person will draw a red marble on the second draw from the bowl without replacing the marble from the first draw, is
A) $\dfrac{2}{3}$
B) $\dfrac{1}{4}$
C) $\dfrac{5}{{12}}$
D) $\dfrac{5}{8}$

Answer 1

Hint : The probability is calculated for both the balls to be chosen in the first draw and for the second draw we calculate it specifically to get red but with consideration of probabilities of respective red and green balls that can be chosen in the first draw.
$Probability(P){\text{ }} = {\text{ }}\dfrac{{Favorable{\text{ }}outcomes(f)}}{{Total{\text{ }}outcomes(T)}}$

Complete step-by-step answer :
Number of red balls = 6
Number of green balls = 3
Total number of balls = 6 + 3
= 9
i) Probability for first draw:
Red balls : Favorable outcomes (f) = 6
Total outcomes (T) = 9
Probability 🡪 $P = \dfrac{f}{T}$
${P_{{R_1}}} = \dfrac{6}{9}$
Green balls : Favorable outcomes (f) = 3
Total outcomes (T) = 9
Probability 🡪 $P = \dfrac{f}{T}$
${P_{{G_1}}} = \dfrac{3}{9}$
ii) Probability for second draw:
As the balls are drawn without replacement, the balls from respective colors and the total number of balls decreases by 1.
a) First ball drawn is red and second drawn is also red
Red marbles = 6 – 1
= 5
Green marbles = 3
Favorable outcomes (f) = 5
Total outcomes (T) = 8
Probability 🡪 $P = \dfrac{f}{T}$
${P_{{R_{2a}}}} = \dfrac{5}{8}$
a) First ball drawn is green and second drawn is red
Red marbles = 6
Green marbles = 3 – 1
= 2
Favorable outcomes (f) = 6
Total outcomes (T) = 8
Probability 🡪 $P = \dfrac{f}{T}$
${P_{{R_{2b}}}} = \dfrac{6}{8}$
Now, the required probability that a red ball is chosen in second draw is given as:
Red in first draw and red in second or green in first draw and red in second –
$P = {P_{{R_1}}} \times {P_{{R_{2a}}}} + {P_{{G_1}}} \times {P_{{R_{2b}}}}$
Substituting the values, we get:

$$P = \dfrac{6}{9} \times \dfrac{5}{8} + \dfrac{3}{9} \times \dfrac{6}{8} \\\ P = \dfrac{{30}}{{72}} + \dfrac{{18}}{{72}} \\\ P = \dfrac{{48}}{{72}} \\\ P = \dfrac{2}{3} \\\$$

Therefore, the probability that a blind folded person will draw a red marble on the second draw from the given bowl without replacing the marble from the first draw, is $\dfrac{2}{3}$ or 0.66.
So, the correct answer is “Option A”.

Note : The number of balls always decreases in case of without replacement because once drawn the ball won’t get mixed again and hence will not be considered for the next draw.
In probability where we use and it means product whereas or means addition
A or B = A X B
A and B = A + B
The probability can either be given in terms of fraction or a decimal