NCERT Solutions · Class 12 Mathematics Probability

Let \(\displaystyle R_1\) and \(\displaystyle B_1\) be the events that the first ball drawn is red and black respectively, and let \(\displaystyle R_2\) be the event that the second ball drawn is red. Since \(\displaystyle R_1, B_1\) are mutually exclusive and exhaustive, the theorem of total probability gives \[P(R_2)=P(R_1)\,P(R_2\mid R_1)+P(B_1)\,P(R_2\mid B_1).\]At the first draw the urn has $\displaystyle 5$ red and $\displaystyle 5$ black balls, so \[P(R_1)=\frac{5}{10}=\frac12,\qquad P(B_1)=\frac{5}{10}=\frac12 .\]The drawn ball is returned and $\displaystyle 2$ balls of its colour are added, so the urn contains \(\displaystyle 10+2=12\) balls at the second draw.If the first ball was red, the urn is $\displaystyle 7$ red and $\displaystyle 5$ black: \[P(R_2\mid R_1)=\frac{7}{12}.\] If the first ball was black, the urn is $\displaystyle 5$ red and $\displaystyle 7$ black: \[P(R_2\mid B_1)=\frac{5}{12}.\]Therefore \[P(R_2)=\frac12\cdot\frac{7}{12}+\frac12\cdot\frac{5}{12}=\frac{7}{24}+\frac{5}{24}=\frac{12}{24}.\]The probability that the second ball is red is \(\displaystyle \dfrac{1}{2}\) — exactly the probability of a red ball at the first draw, because the urn was symmetric in the two colours to begin with.

Let \(\displaystyle E_1\): the first bag is chosen, \(\displaystyle E_2\): the second bag is chosen, and \(\displaystyle A\): the ball drawn is red. We need \(\displaystyle P(E_1\mid A)\), so this is Bayes' theorem: \[P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}.\]A bag is selected at random, so \[P(E_1)=P(E_2)=\frac12 .\]Bag $\displaystyle 1$ has $\displaystyle 4$ red out of \(\displaystyle 4+4=8\) balls and bag $\displaystyle 2$ has $\displaystyle 2$ red out of \(\displaystyle 2+6=8\) balls: \[P(A\mid E_1)=\frac{4}{8}=\frac12,\qquad P(A\mid E_2)=\frac{2}{8}=\frac14 .\]Substituting, \[P(E_1\mid A)=\frac{\frac12\cdot\frac12}{\frac12\cdot\frac12+\frac12\cdot\frac14} =\frac{\frac14}{\frac14+\frac18}=\frac{\frac14}{\frac38}=\frac14\cdot\frac83 .\]The probability that the red ball came from the first bag is \(\displaystyle \dfrac{2}{3}\).

Let \(\displaystyle E_1\): the student is a hostlier, \(\displaystyle E_2\): the student is a day scholar, and \(\displaystyle A\): the student attains A grade. From the data, \[P(E_1)=60\%=\frac{60}{100}=0.6,\qquad P(E_2)=40\%=0.4,\] \[P(A\mid E_1)=30\%=0.3,\qquad P(A\mid E_2)=20\%=0.2 .\] (Note the two 'A grade' percentages are conditional rates within each group, not shares of the whole college.)By Bayes' theorem, \[P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)} =\frac{0.6\times 0.3}{0.6\times 0.3+0.4\times 0.2}.\]\[P(E_1\mid A)=\frac{0.18}{0.18+0.08}=\frac{0.18}{0.26}=\frac{18}{26}.\]The probability that a randomly chosen A-grade student is a hostlier is \(\displaystyle \dfrac{9}{13}\).

Let \(\displaystyle E_1\): the student knows the answer, \(\displaystyle E_2\): the student guesses, and \(\displaystyle A\): the answer given is correct. Given \[P(E_1)=\frac34,\qquad P(E_2)=\frac14,\qquad P(A\mid E_2)=\frac14 .\]The step to be careful about: if the student knows the answer he is certainly correct, so \[P(A\mid E_1)=1 .\]By Bayes' theorem, \[P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)} =\frac{\frac34\times 1}{\frac34\times 1+\frac14\times\frac14}.\]\[P(E_1\mid A)=\frac{\frac34}{\frac34+\frac1{16}}=\frac{\dfrac{12}{16}}{\dfrac{13}{16}}.\]The probability that the student actually knew the answer, given that he answered correctly, is \(\displaystyle \dfrac{12}{13}\).

Let \(\displaystyle E_1\): the person has the disease, \(\displaystyle E_2\): the person is healthy, and \(\displaystyle A\): the test result is positive.The prevalence is $\displaystyle 0.1$ percent, i.e. \(\displaystyle 0.1\% = \dfrac{0.1}{100}\): \[P(E_1)=0.001,\qquad P(E_2)=1-0.001=0.999 .\] (Writing \(\displaystyle P(E_1)=0.1\) here is the usual slip — $\displaystyle 0.1$ per cent is 0.001.)The test is $\displaystyle 99$% effective on the diseased and gives a false positive with probability $\displaystyle 0.005$: \[P(A\mid E_1)=0.99,\qquad P(A\mid E_2)=0.005 .\]By Bayes' theorem, \[P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)} =\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}.\]\[P(E_1\mid A)=\frac{0.00099}{0.00099+0.004995}=\frac{0.00099}{0.005985}=\frac{990}{5985}=\frac{22}{133}.\]The probability that a person with a positive test actually has the disease is \(\displaystyle \dfrac{22}{133}\approx 0.165\) — barely one positive in six is genuine, because the disease is so rare that false positives from the huge healthy group outnumber the true ones.

Let \(\displaystyle E_1\): the two-headed coin is chosen, \(\displaystyle E_2\): the biased coin is chosen, \(\displaystyle E_3\): the unbiased coin is chosen, and \(\displaystyle A\): the toss shows a head.One of the three coins is chosen at random, so \[P(E_1)=P(E_2)=P(E_3)=\frac13 .\]The conditional probabilities of a head are \[P(A\mid E_1)=1\ \ (\text{both faces are heads}),\qquad P(A\mid E_2)=\frac{75}{100}=\frac34,\qquad P(A\mid E_3)=\frac12 .\]By Bayes' theorem, \[P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)} =\frac{\frac13\times 1}{\frac13\times 1+\frac13\times\frac34+\frac13\times\frac12}.\]The common factor \(\displaystyle \frac13\) cancels: \[P(E_1\mid A)=\frac{1}{1+\frac34+\frac12}=\frac{1}{\dfrac{4+3+2}{4}}=\frac{4}{9}.\]The probability that the coin tossed was the two-headed coin is \(\displaystyle \dfrac{4}{9}\).

Let \(\displaystyle E_1, E_2, E_3\) be the events that the insured person is a scooter driver, a car driver and a truck driver respectively, and \(\displaystyle A\) the event that he meets with an accident.Total number insured \(\displaystyle =2000+4000+6000=12000\), so \[P(E_1)=\frac{2000}{12000}=\frac16,\qquad P(E_2)=\frac{4000}{12000}=\frac13,\qquad P(E_3)=\frac{6000}{12000}=\frac12 .\]The accident rates are \[P(A\mid E_1)=0.01=\frac{1}{100},\qquad P(A\mid E_2)=0.03=\frac{3}{100},\qquad P(A\mid E_3)=0.15=\frac{15}{100}.\]By Bayes' theorem, \[P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}.\]Numerator \(\displaystyle =\dfrac16\times\dfrac{1}{100}=\dfrac{1}{600}\).Denominator \(\displaystyle =\dfrac16\cdot\dfrac{1}{100}+\dfrac13\cdot\dfrac{3}{100}+\dfrac12\cdot\dfrac{15}{100} =\dfrac{1}{600}+\dfrac{1}{100}+\dfrac{15}{200}=\dfrac{1+6+45}{600}=\dfrac{52}{600}\).Hence \[P(E_1\mid A)=\frac{1/600}{52/600}=\frac{1}{52}.\]The probability that the person who met with an accident is a scooter driver is \(\displaystyle \dfrac{1}{52}\).

Let \(\displaystyle E_1\): the item was produced by machine A, \(\displaystyle E_2\): by machine B, and \(\displaystyle A\): the item chosen is defective. Given \[P(E_1)=60\%=0.6,\qquad P(E_2)=40\%=0.4,\] \[P(A\mid E_1)=2\%=0.02,\qquad P(A\mid E_2)=1\%=0.01 .\]The question asks for \(\displaystyle P(E_2\mid A)\) (machine B), so by Bayes' theorem \[P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)} =\frac{0.4\times 0.01}{0.6\times 0.02+0.4\times 0.01}.\]\[P(E_2\mid A)=\frac{0.004}{0.012+0.004}=\frac{0.004}{0.016}.\]The probability that the defective item was produced by machine B is \(\displaystyle \dfrac{1}{4}\).

Let \(\displaystyle E_1\): the first group wins, \(\displaystyle E_2\): the second group wins, and \(\displaystyle A\): a new product is introduced. Given \[P(E_1)=0.6,\qquad P(E_2)=0.4,\qquad P(A\mid E_1)=0.7,\qquad P(A\mid E_2)=0.3 .\]We need the probability that the second group was responsible, given that the new product was introduced, i.e. \(\displaystyle P(E_2\mid A)\). By Bayes' theorem, \[P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)} =\frac{0.4\times 0.3}{0.6\times 0.7+0.4\times 0.3}.\]\[P(E_2\mid A)=\frac{0.12}{0.42+0.12}=\frac{0.12}{0.54}=\frac{12}{54}.\]The probability that the new product was introduced by the second group is \(\displaystyle \dfrac{2}{9}\).

Let \(\displaystyle E_1\): the die shows $\displaystyle 5$ or $\displaystyle 6$, \(\displaystyle E_2\): the die shows $\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$ or $\displaystyle 4$, and \(\displaystyle A\): exactly one head is obtained. Then \[P(E_1)=\frac{2}{6}=\frac13,\qquad P(E_2)=\frac{4}{6}=\frac23 .\]If \(\displaystyle E_1\) occurs the coin is tossed three times, so by the binomial distribution with \(\displaystyle n=3,\ p=\frac12\), \[P(A\mid E_1)=\binom{3}{1}\left(\frac12\right)^{1}\left(\frac12\right)^{2}=\frac38 .\] (Equivalently HTT, THT, TTH out of $\displaystyle 8$ equally likely outcomes.)If \(\displaystyle E_2\) occurs the coin is tossed only once, and 'exactly one head' means the single toss is a head: \[P(A\mid E_2)=\frac12 .\]By Bayes' theorem, \[P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)} =\frac{\frac23\cdot\frac12}{\frac13\cdot\frac38+\frac23\cdot\frac12} =\frac{\frac13}{\frac18+\frac13}.\]\[P(E_2\mid A)=\frac{\frac13}{\dfrac{3+8}{24}}=\frac13\cdot\frac{24}{11}=\frac{8}{11}.\]The probability that she threw $\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$ or $\displaystyle 4$ with the die is \(\displaystyle \dfrac{8}{11}\).