A bag contains 5 balls of unknown colour. A ball is drawn and replaced twice , and in each case it is found to be red.: if two balls are drawn simultaneously .find the chances that it is red.
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2 Answers
1It's odd that i'm posting this question's solution as i found it's solution finally.
RED | NOT RED
cse 1 2 | 3 pP1=2/5 .2/5 P= 4/25 Q1 = 4/25/11/5 = 4/55
case 2 3 | 2 pP2=3/5 . 3/5P = 9/25 Q2 = 9/25/11/25 = 9/55
case 3 4 | 1 pP3=4/5 . 4/5 P= 16/25 similarly Q3 = 16/55
case 4 5 | 0 pP4=1.1 = 1P Q4 = 5/11
case 1 | 4 pP5=1/5 . 1/5P = 1/25 Q5 = 1/55
\sum{p_{r}P_{r}} = (4/25 + 1/25 + 9/25 + 16/25 + 1 )P= \frac{11}{5}P
for both red from
case 1 p = 2/5 . 1/4 . 4/55
case 2 p = 3/5 .1/2 . 9/55
case 3 p = 4/5.3/2.16/55
case 4 p = 1 . 1 . 5/11
case 5 p = 0 . 1/55
adding all the abv gives required probabilty.