question_answer

A bag contains 25 balls of which 10 are purple and the remaining are pink. A ball is drawn at random, its colour is noted and it is replaced. 6 balls are drawn in this way. Find the probability that (i) All balls were purple, (ii) Not more than 2 were pink. (iii) An equal number of purple and pink balls were drawn. (iv) atleast one ball was pink.

Answer:

The given question is the case of Bernoulli?s trials. Let success : Getting a purple ball on a draw Failure: Getting  a pink ball on a draw \[P=P\,(success)=\frac{10}{25}=\frac{2}{5}\Rightarrow q=\frac{3}{5}\] (i) P (6 success) \[={}^{6}{{C}_{6}}{{p}^{6}}{{q}^{0}}=1\times {{\left( \frac{2}{5} \right)}^{6}}\times 1={{\left( \frac{2}{5} \right)}^{6}}\] (ii) P (not more than 2 failure) = P (not less then 4 success) \[=P(4)+P(5)+P(6)\] \[={}^{6}{{C}_{4}}{{P}^{4}}{{q}^{2}}+{}^{6}{{C}_{5}}{{P}^{5}}{{q}^{1}}+{}^{6}{{C}_{6}}{{P}^{6}}{{q}^{0}}\] \[=15\times {{\left( \frac{2}{5} \right)}^{4}}{{\left( \frac{3}{5} \right)}^{2}}+6\times {{\left( \frac{2}{5} \right)}^{5}}\left( \frac{3}{5} \right)+1\times {{\left( \frac{2}{5} \right)}^{6}}\] \[={{\left( \frac{2}{5} \right)}^{4}}\left[ 15\times {{\left( \frac{3}{5} \right)}^{2}}+6\left( \frac{2}{5} \right)\left( \frac{3}{5} \right)+{{\left( \frac{2}{5} \right)}^{2}} \right]\] \[={{\left( \frac{2}{5} \right)}^{4}}\left[ \frac{135}{25}+\frac{36}{25}+\frac{4}{25} \right]\] \[={{\left( \frac{2}{5} \right)}^{4}}\left[ \frac{175}{25} \right]=7\times {{\left( \frac{2}{5} \right)}^{4}}\] (iii) P (3 success 7 failure) = P (3) \[={}^{6}{{C}_{3}}{{p}^{3}}{{q}^{3}}=15{{\left( \frac{2}{5} \right)}^{3}}{{\left( \frac{3}{5} \right)}^{3}}=\frac{864}{3125}\] (iv) P (atleast 1 failure) = P (at most 5 success) \[=P\,(0)+P\,(1)+P\,(2)+P\,(3)+P\,(4)+P\,(5)\] \[=1-P(6)=1-{}^{6}{{C}_{6}}{{p}^{6}}{{q}^{0}}\] \[=1-1\times {{\left( \frac{2}{5} \right)}^{6}}\times 1=1-{{\left( \frac{2}{5} \right)}^{6}}\]