Question

Bag A contains 3 red and 3 blue balls.Bag B contains 1 red and 3 blue balls.A ball is taken at random from bag A and placed in bag B. What is the probability that the ball taken from B is red? [ solve without using bayes theorem]​

Answer 1

Answer:

401/1890

Step-by-step explanation:

Probability of selecting a bag out of three =  1/3

Probability of getting a red and white ball from bag A

=> ${1}_ C_{1} X {3} _C_{1} / {6}_C_{2} = 3/15 = 1/5$

Probability of selecting a red and a white ball from bag B

=> ${2}_C_{1} X {3}_C_{1} / {8}_C_{2} = 6/28= 3/14$

Probability of selecting a red and a white ball from bag C

=> ${4}_C_{1} X {2}_C_{1} / {9}_C_{2} = 8/36 = 2/9$

So the probability that balls are white and red

=> 1/3(1/5+3/14+2/9)

=> 1/3(126+135+140/630)

=> 401/1890

HENCE PROVED ;)

Please mark it the brainliest

Answer 2

Answer.

401/1890

401/1890Step-by-step explanation:

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3Probability of getting a red and white ball from bag A

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3Probability of getting a red and white ball from bag A=> {1}_ C_{1} X {3} _C_{1} / {6}_C_{2} = 3/15 = 1/5

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3Probability of getting a red and white ball from bag A=> {1}_ C_{1} X {3} _C_{1} / {6}_C_{2} = 3/15 = 1/5Probability of selecting a red and a white ball from bag B

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3Probability of getting a red and white ball from bag A=> {1}_ C_{1} X {3} _C_{1} / {6}_C_{2} = 3/15 = 1/5Probability of selecting a red and a white ball from bag B=> {2}_C_{1} X {3}_C_{1} / {8}_C_{2} = 6/28= 3/14

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3Probability of getting a red and white ball from bag A=> {1}_ C_{1} X {3} _C_{1} / {6}_C_{2} = 3/15 = 1/5Probability of selecting a red and a white ball from bag B=> {2}_C_{1} X {3}_C_{1} / {8}_C_{2} = 6/28= 3/14Probability of selecting a red and a white ball from bag C

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3Probability of getting a red and white ball from bag A=> {1}_ C_{1} X {3} _C_{1} / {6}_C_{2} = 3/15 = 1/5Probability of selecting a red and a white ball from bag B=> {2}_C_{1} X {3}_C_{1} / {8}_C_{2} = 6/28= 3/14Probability of selecting a red and a white ball from bag C=> {4}_C_{1} X {2}_C_{1} / {9}_C_{2} = 8/36 = 2/9

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3Probability of getting a red and white ball from bag A=> {1}_ C_{1} X {3} _C_{1} / {6}_C_{2} = 3/15 = 1/5Probability of selecting a red and a white ball from bag B=> {2}_C_{1} X {3}_C_{1} / {8}_C_{2} = 6/28= 3/14Probability of selecting a red and a white ball from bag C=> {4}_C_{1} X {2}_C_{1} / {9}_C_{2} = 8/36 = 2/9So the probability that balls are white and red

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3Probability of getting a red and white ball from bag A=> {1}_ C_{1} X {3} _C_{1} / {6}_C_{2} = 3/15 = 1/5Probability of selecting a red and a white ball from bag B=> {2}_C_{1} X {3}_C_{1} / {8}_C_{2} = 6/28= 3/14Probability of selecting a red and a white ball from bag C=> {4}_C_{1} X {2}_C_{1} / {9}_C_{2} = 8/36 = 2/9So the probability that balls are white and red=> 1/3(1/5+3/14+2/9)

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3Probability of getting a red and white ball from bag A=> {1}_ C_{1} X {3} _C_{1} / {6}_C_{2} = 3/15 = 1/5Probability of selecting a red and a white ball from bag B=> {2}_C_{1} X {3}_C_{1} / {8}_C_{2} = 6/28= 3/14Probability of selecting a red and a white ball from bag C=> {4}_C_{1} X {2}_C_{1} / {9}_C_{2} = 8/36 = 2/9So the probability that balls are white and red=> 1/3(1/5+3/14+2/9)=> 1/3(126+135+140/630)

401/1890Step-by-step explanation:Probability of selecting a bag out of three = 1/3Probability of getting a red and white ball from bag A=> {1}_ C_{1} X {3} _C_{1} / {6}_C_{2} = 3/15 = 1/5Probability of selecting a red and a white ball from bag B=> {2}_C_{1} X {3}_C_{1} / {8}_C_{2} = 6/28= 3/14Probability of selecting a red and a white ball from bag C=> {4}_C_{1} X {2}_C_{1} / {9}_C_{2} = 8/36 = 2/9So the probability that balls are white and red=> 1/3(1/5+3/14+2/9)=> 1/3(126+135+140/630)=> 401/1890

Hence Proved.