Question

A bag contains 5 balls of unknown colour; a ball is drawn and replaced twice, and in each case is found to be red: if two balls are now drawn simultaneously find the chance that both are red.

Answer 1

Hint: In order to solve this problem find the cases that at least one of the balls is red then find the probability of red balls out of five balls and then see which of the probabilities satisfies the conditions asked.

Complete step-by-step answer:
The 5 likely cases: 1) All 5 balls are red, 2) 4 balls are red, 3) 3 balls are red, 4) 2 balls are red, 5) 1 ball is red

Probability of drawing red balls in two repeated drawings when all 5 balls are red = ${\left( {\dfrac{5}{5}} \right)^2} = 1$
Probability of drawing red balls in two repeated drawings when 4 balls are red = ${\left( {\dfrac{4}{5}} \right)^2}$
Probability of drawing red balls in two repeated drawings when 3 balls are red = ${\left( {\dfrac{3}{5}} \right)^2}$
Probability of drawing red balls in two repeated drawings when 2 balls are red = ${\left( {\dfrac{2}{5}} \right)^2}$
Probability of drawing red balls in two repeated drawings when 1 ball is red = ${\left( {\dfrac{1}{5}} \right)^2}$
 Probability of 5 balls to be red = $\dfrac{{{{\left( {\dfrac{5}{5}} \right)}^2}}}{{{{\left( {\dfrac{5}{5}} \right)}^2} + {{\left( {\dfrac{4}{5}} \right)}^2} + {{\left( {\dfrac{3}{5}} \right)}^2} + {{\left( {\dfrac{2}{5}} \right)}^2} + {{\left( {\dfrac{1}{5}} \right)}^2}}} = \dfrac{5}{{11}}$
Similarly, Probability of 4 balls to be red = $\dfrac{{{{\left( {\dfrac{4}{5}} \right)}^2}}}{{{{\left( {\dfrac{5}{5}} \right)}^2} + {{\left( {\dfrac{4}{5}} \right)}^2} + {{\left( {\dfrac{3}{5}} \right)}^2} + {{\left( {\dfrac{2}{5}} \right)}^2} + {{\left( {\dfrac{1}{5}} \right)}^2}}} = \dfrac{{16}}{{55}}$
Similarly, Probability of 3 balls to be red = $\dfrac{{{{\left( {\dfrac{3}{5}} \right)}^2}}}{{{{\left( {\dfrac{5}{5}} \right)}^2} + {{\left( {\dfrac{4}{5}} \right)}^2} + {{\left( {\dfrac{3}{5}} \right)}^2} + {{\left( {\dfrac{2}{5}} \right)}^2} + {{\left( {\dfrac{1}{5}} \right)}^2}}} = \dfrac{9}{{55}}$
Similarly, Probability of 2 balls to be red = $\dfrac{{{{\left( {\dfrac{2}{5}} \right)}^2}}}{{{{\left( {\dfrac{5}{5}} \right)}^2} + {{\left( {\dfrac{4}{5}} \right)}^2} + {{\left( {\dfrac{3}{5}} \right)}^2} + {{\left( {\dfrac{2}{5}} \right)}^2} + {{\left( {\dfrac{1}{5}} \right)}^2}}} = \dfrac{4}{{55}}$
Similarly, Probability of 1 ball to be red = $\dfrac{{{{\left( {\dfrac{1}{5}} \right)}^2}}}{{{{\left( {\dfrac{5}{5}} \right)}^2} + {{\left( {\dfrac{4}{5}} \right)}^2} + {{\left( {\dfrac{3}{5}} \right)}^2} + {{\left( {\dfrac{2}{5}} \right)}^2} + {{\left( {\dfrac{1}{5}} \right)}^2}}} = \dfrac{1}{{55}}$
Therefore the probability of drawing 2 red balls simultaneously
= $\dfrac{1}{{55}} \times 0 + \dfrac{4}{{55}} \times \dfrac{1}{{10}} + \dfrac{9}{{55}} \times \dfrac{3}{{10}} + \dfrac{{16}}{{55}} \times \dfrac{6}{{10}} + \dfrac{5}{{11}} \times \dfrac{{10}}{{10}} = \dfrac{{377}}{{550}}$
Therefore the answer to this problem is $\dfrac{{377}}{{550}}$.

Note: When you get to solve such problems of probability then you need to consider all the possibilities that come under required conditions and solve according to that condition provided in the problem. Using probability is equal to the number of favourable outcomes upon total number of outcomes is the basic and all time formula needed to calculate the probability.