question_answer

It is known that 10% of certain articles manufactured are defective. What is probability that in a random sample such articles, 9 are defective?

OR

Consider the experiment of tossing a coin. If the coin shows tail, toss it again but if it shows head, then throw a die. Find the conditional probability off event that 'the die shows a number greater than 3' given that 'there is at least one head'.

Answer:

As, 10% of certain articles manufactured are defective. We need to determine the probability that in a random sample of (n = 12) such articles, 9 are defective. Let x be the number of times we select a defective article out of 12 articles. It is a Bernoulli trial as they satisfy the conditions (i) finite numbers of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial. P (getting a defective article)\[p=10%=\frac{1}{10}\] and                   \[q=1-P=\frac{q}{10}\]. Since X has a binomial distribution, the probability of success in n-Bernoulli trials,\[P\,(x=x)={}^{c}{{C}_{x}}{{P}^{x}}{{q}^{n-x}}\]where \[x=0,\,\,1,\,\,2,\,\,....,\,\,n\]and \[q=(1-p)\]. We need to calculate the probability of 9 items being defective, i.e. \[P\,(X=9)={}^{12}{{C}_{p}}{{\left( \frac{1}{10} \right)}^{9}}{{\left( \frac{9}{10} \right)}^{12-9}}\] \[=\frac{12\times 11\times 10\times 9!}{9!\times 3\times 2\times 1}{{\left( \frac{1}{10} \right)}^{9}}{{\left( \frac{9}{10} \right)}^{3}}\] \[=22\,\left( \frac{{{9}^{3}}}{{{10}^{11}}} \right)\] Head and 4 probability\[=\frac{1}{2}\times \frac{1}{6}\] Head and 5 probability\[=\frac{1}{2}\times \frac{1}{6}\] Head and 6 probability\[=\frac{1}{2}\times \frac{1}{6}\] Favourable probability\[=\frac{1}{4}\] At least/Head probability\[=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\] \[\therefore \]Conditional probability\[=\frac{1}{4}\div \frac{3}{4}=\frac{1}{3}\]