The probabilities of two students A and B coming to the school in time are \[\frac{3}{7}\] and \[\frac{5}{7},\] respectively. Assuming that the events, ?A coming in time' and 'B coming in time' are independent, find the probability, of only one of them coming to the school in time |
OR |
Find the probability that in 10 throws of a fair die, a score which is a multiple of 3 will be obtained in at least 8 of the throws. |
Answer:
P(A) = Probability of student A coming to school in time \[=\frac{3}{7}\] P(B) = Probability of student B coming to school in time \[=\frac{5}{7}\] Then, the probability that only one of students coming to school in time \[=P(A\,\cap \bar{B})+P(\bar{A}\,\cap B)\] \[=P(A)\times P(\bar{B})+P(\bar{A})\times P(B)\] \[=P(A)\times [1-P(B)]+[1-P(A)]\times P(B)\] \[=\frac{3}{7}\times \left( 1-\frac{5}{7} \right)+\left( 1-\frac{3}{7} \right)\times \frac{5}{7}\] \[=\frac{3}{7}\times \frac{2}{7}+\frac{4}{7}\times \frac{5}{7}\] \[=\frac{6}{49}+\frac{20}{49}=\frac{26}{49}\] OR Here, success is a score which is a multiple of 3, i.e. 3 or 6. Let p = probability of success \[=\frac{2}{6}=\frac{1}{3}\] and q = probability of failure \[=1-\frac{1}{3}=\frac{2}{3}\] Now, let X be a random variable that denotes the number of success in 10 throws of a fair die. Then, X can take values 0, 1, 2,...,10. Clearly, X follows binomial distribution with n = 10 and \[p=\frac{1}{3}.\] \[\therefore P(X=x){{=}^{10}}{{C}_{x}}{{\left( \frac{1}{3} \right)}^{x}}{{\left( \frac{2}{3} \right)}^{10-x}};x=0,\,\,1,\,\,2,...10\] Now, required probability \[=P(X\ge 8)\] \[=P(X=8)+P(X=9)+P(X=10)\] \[{{=}^{10}}{{C}_{8}}{{\left( \frac{1}{3} \right)}^{8}}{{\left( \frac{2}{3} \right)}^{2}}{{+}^{10}}{{C}_{9}}{{\left( \frac{1}{3} \right)}^{9}}\left( \frac{2}{3} \right){{+}^{10}}{{C}_{10}}{{\left( \frac{1}{3} \right)}^{10}}\] \[={{\left( \frac{1}{3} \right)}^{10}}{{[}^{10}}{{C}_{2}}.4\,{{+}^{10}}{{C}_{1}}\cdot 2+1]\] \[={{\left( \frac{1}{3} \right)}^{10}}[45\times 4+20+1]=\frac{201}{{{3}^{10}}}\]
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