Quantity B: Probability that the ball is from the first bag if the dra

1.	Quantity B: Probability that the ball is from the first bag if the drawn ball is of blue colour.1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B
Answer» ⇒ Probability of selection of a ball from bag 1 = P(B₁) = 1/2 ⇒ Probability of selection of a ball from bag 2 = P(B₂) = 1/2 Now, there are total 11 balls in the bag 1 (4 red balls + 7 blue balls), ⇒ Probability of getting red ball from bag 1 = P(R/B₁) = number of red balls/total balls = 4/11 ⇒ Probability of getting blue ball from bag 1 = P(B/B₁) = number of blue balls/total balls = 7/11 Also, there are total 8 balls in the bag 2 (5 red balls + 3 blue balls), ⇒ Probability of getting red ball from bag 2 = P(R/B₂) = number of red balls/total balls = 5/8 ⇒ Probability of getting blue ball from bag 2 = P(B/B₂) = number of blue balls/total balls = 3/8 Quantity A: It is given that ball DRAWN is red, we have to find the probability of drawn ball being from bag 2, So, using Baye’s theorem, ⇒ P(B₂/R) = {P(B₂) × P(R/B₂)}/[{P(B₂) × P(R/B₂)} + {P(B₁) × P(R/B₁)}] ⇒ P(B₂/R) = {(1/2) × (5/8)} / [{(1/2) × (5/8)} + {(1/2) × (4/11) $(\Rightarrow {\rm{\;P}}\left( {{{\rm{B}}_2}/{\rm{R}}} \right) = \FRAC{{{\rm{P}}\left( {{{\rm{B}}_2}} \right) \times {\rm{P}}\left( {\frac{{\rm{R}}}{{{{\rm{B}}_2}}}} \right)}}{{{\rm{P}}\left( {{{\rm{B}}_2}} \right) \times {\rm{P}}\left( {\frac{{\rm{R}}}{{{{\rm{B}}_2}}}} \right) + \;{\rm{P}}\left( {{{\rm{B}}_1}} \right) \times {\rm{P}}\left( {\frac{{\rm{R}}}{{{{\rm{B}}_1}}}} \right)}})$ Putting all the values in the formula, we get $(\Rightarrow {\rm{P}}\left( {{{\rm{B}}_2}/{\rm{R}}} \right) = {\rm{\;}}\frac{{\frac{1}{{2{\rm{\;}}}} \times \frac{5}{8}}}{{\frac{1}{2} \times \frac{5}{8}\; + \;\frac{1}{2} \times \frac{4}{{11}}}} = \frac{{\frac{5}{{16}}}}{{\frac{5}{{16}}\; + \;\frac{4}{{22}}}} = 110/174{\rm{\;}} = {\rm{\;}}55/87)$ ⇒ Quantity A = 55/87 Quantity B: It is given that ball drawn is blue, we have to find the probability of drawn ball being from bag 1, So, using Baye’s theorem, $(\Rightarrow {\rm{\;P}}\left( {\frac{{{{\rm{B}}_1}}}{{\rm{B}}}} \right) = \frac{{{\rm{P}}\left( {{{\rm{B}}_1}} \right) \times {\rm{\;P}}\left( {\frac{{\rm{B}}}{{{{\rm{B}}_1}}}} \right)}}{{{\rm{P}}\left( {{{\rm{B}}_1}} \right) \times {\rm{\;P}}\left( {\frac{{\rm{B}}}{{{{\rm{B}}_1}}}} \right) + \;{\rm{P}}\left( {{{\rm{B}}_2}} \right) \times {\rm{\;P}}\left( {\frac{{\rm{B}}}{{{{\rm{B}}_2}}}} \right)}})$ Putting all the values in the formula, $(\Rightarrow {\rm{\;P}}\left( {\frac{{{{\rm{B}}_1}}}{{\rm{B}}}} \right) = {\rm{\;}}\frac{{\frac{1}{2} \times \frac{7}{{11}}}}{{\frac{1}{2} \times \frac{7}{{11}}\; + \;\frac{1}{2} \times \frac{3}{8}}} = \frac{{\frac{7}{{22}}}}{{\frac{7}{{22}}\; + \;\frac{3}{{16}}}}{\rm{\;}} = 112/178{\rm{\;}} = {\rm{\;}}56/89)$ ⇒ Quantity B = 56/89 ∴ Quantity A > Quantity B

Quantity B: Probability that the ball is from the first bag if the drawn ball is of blue colour.1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

Answer»

⇒ Probability of selection of a ball from bag 1 = P(B₁) = 1/2

⇒ Probability of selection of a ball from bag 2 = P(B₂) = 1/2

Now, there are total 11 balls in the bag 1 (4 red balls + 7 blue balls),

⇒ Probability of getting red ball from bag 1 = P(R/B₁) = number of red balls/total balls = 4/11

⇒ Probability of getting blue ball from bag 1 = P(B/B₁) = number of blue balls/total balls = 7/11

Also, there are total 8 balls in the bag 2 (5 red balls + 3 blue balls),

⇒ Probability of getting red ball from bag 2 = P(R/B₂) = number of red balls/total balls = 5/8

⇒ Probability of getting blue ball from bag 2 = P(B/B₂) = number of blue balls/total balls = 3/8

Quantity A:

It is given that ball DRAWN is red, we have to find the probability of drawn ball being from bag 2,

So, using Baye’s theorem,

⇒ P(B₂/R) = {P(B₂) × P(R/B₂)}/[{P(B₂) × P(R/B₂)} + {P(B₁) × P(R/B₁)}]

⇒ P(B₂/R) = {(1/2) × (5/8)} / [{(1/2) × (5/8)} + {(1/2) × (4/11)

$(\Rightarrow {\rm{\;P}}\left( {{{\rm{B}}_2}/{\rm{R}}} \right) = \FRAC{{{\rm{P}}\left( {{{\rm{B}}_2}} \right) \times {\rm{P}}\left( {\frac{{\rm{R}}}{{{{\rm{B}}_2}}}} \right)}}{{{\rm{P}}\left( {{{\rm{B}}_2}} \right) \times {\rm{P}}\left( {\frac{{\rm{R}}}{{{{\rm{B}}_2}}}} \right) + \;{\rm{P}}\left( {{{\rm{B}}_1}} \right) \times {\rm{P}}\left( {\frac{{\rm{R}}}{{{{\rm{B}}_1}}}} \right)}})$

Putting all the values in the formula, we get

$(\Rightarrow {\rm{P}}\left( {{{\rm{B}}_2}/{\rm{R}}} \right) = {\rm{\;}}\frac{{\frac{1}{{2{\rm{\;}}}} \times \frac{5}{8}}}{{\frac{1}{2} \times \frac{5}{8}\; + \;\frac{1}{2} \times \frac{4}{{11}}}} = \frac{{\frac{5}{{16}}}}{{\frac{5}{{16}}\; + \;\frac{4}{{22}}}} = 110/174{\rm{\;}} = {\rm{\;}}55/87)$

⇒ Quantity A = 55/87

Quantity B:

It is given that ball drawn is blue, we have to find the probability of drawn ball being from bag 1,

So, using Baye’s theorem,

$(\Rightarrow {\rm{\;P}}\left( {\frac{{{{\rm{B}}_1}}}{{\rm{B}}}} \right) = \frac{{{\rm{P}}\left( {{{\rm{B}}_1}} \right) \times {\rm{\;P}}\left( {\frac{{\rm{B}}}{{{{\rm{B}}_1}}}} \right)}}{{{\rm{P}}\left( {{{\rm{B}}_1}} \right) \times {\rm{\;P}}\left( {\frac{{\rm{B}}}{{{{\rm{B}}_1}}}} \right) + \;{\rm{P}}\left( {{{\rm{B}}_2}} \right) \times {\rm{\;P}}\left( {\frac{{\rm{B}}}{{{{\rm{B}}_2}}}} \right)}})$

Putting all the values in the formula,

$(\Rightarrow {\rm{\;P}}\left( {\frac{{{{\rm{B}}_1}}}{{\rm{B}}}} \right) = {\rm{\;}}\frac{{\frac{1}{2} \times \frac{7}{{11}}}}{{\frac{1}{2} \times \frac{7}{{11}}\; + \;\frac{1}{2} \times \frac{3}{8}}} = \frac{{\frac{7}{{22}}}}{{\frac{7}{{22}}\; + \;\frac{3}{{16}}}}{\rm{\;}} = 112/178{\rm{\;}} = {\rm{\;}}56/89)$

Quantity B: Probability that the ball is from the first bag if the drawn ball is of blue colour.1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

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