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

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) 2/3 (B) 1/4 (C) 5/12 (D) 5/8

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/44daa6b4d8640bd9f00d2b5d7d3e39b9-.png" /> E: Event that the 2 text nd drawn marble is red; R: 1 text st drawn is red; G=1 text st drawn is green P(E) =P(R ∩ E)+P(G ∩ E) =P(R) ⋅ P(E / R)+P(G) ⋅ P(E / G) =(6/9) ⋅ (5/8)+(3/9) ⋅ (6/8)=(48/72)=(2/3)

Q. 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

2/3

1/4

5/12

5/8

$E$ : Event that the $2^{nd}$ drawn marble is red; $R : 1^{st}$ drawn is red; $G = 1^{st}$ drawn is green
$P (E) = P (R \cap E) + P (G \cap E)$
$= P (R) \cdot P (E / R) + P (G) \cdot P (E / G)$
$= \frac{6}{9} \cdot \frac{5}{8} + \frac{3}{9} \cdot \frac{6}{8} = \frac{48}{72} = \frac{2}{3}$

Q. 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

2/3

1/4

5/12

5/8

E : Event that the 2nd drawn marble is red; R:1st drawn is red; G=1st drawn is green P(E)=P(R∩E)+P(G∩E) =P(R)⋅P(E/R)+P(G)⋅P(E/G) =96​⋅85​+93​⋅86​=7248​=32​

$E$ : Event that the $2^{nd}$ drawn marble is red; $R : 1^{st}$ drawn is red; $G = 1^{st}$ drawn is green
$P (E) = P (R \cap E) + P (G \cap E)$
$= P (R) \cdot P (E / R) + P (G) \cdot P (E / G)$
$= \frac{6}{9} \cdot \frac{5}{8} + \frac{3}{9} \cdot \frac{6}{8} = \frac{48}{72} = \frac{2}{3}$