NCERT Solutions · Class 12 Mathematics Probability

By the definition of independent events, A and B are independent precisely when the multiplication rule collapses to a product of the unconditional probabilities: \[\mathrm{P}(\mathrm{A}\cap\mathrm{B})=\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B}). \] Here \(\displaystyle \mathrm{P}(\mathrm{A})=\dfrac{3}{5}\) and \(\displaystyle \mathrm{P}(\mathrm{B})=\dfrac{1}{5}\), and we are told the events are independent, so \[\mathrm{P}(\mathrm{A}\cap\mathrm{B})=\frac{3}{5}\times\frac{1}{5}=\frac{3}{25}. \] \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})=\dfrac{3}{25}\).

Let \(\displaystyle \mathrm{A}\) be the event 'first card is black' and \(\displaystyle \mathrm{B}\) the event 'second card is black'. The draws are without replacement, so the two events are not independent and we must use the multiplication theorem of probability: \[\mathrm{P}(\mathrm{A}\cap\mathrm{B})=\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B}\mid\mathrm{A}). \] A pack of $\displaystyle 52$ cards has $\displaystyle 26$ black cards, so \[\mathrm{P}(\mathrm{A})=\frac{26}{52}=\frac{1}{2}. \] Given that a black card has already been removed, $\displaystyle 25$ black cards remain among the $\displaystyle 51$ cards left — this is the step to get right; the denominator changes too: \[\mathrm{P}(\mathrm{B}\mid\mathrm{A})=\frac{25}{51}. \] Hence \[\mathrm{P}(\text{both black})=\frac{1}{2}\times\frac{25}{51}=\frac{25}{102}. \] The required probability is \(\displaystyle \dfrac{25}{102}\).

The box is approved only if all three drawn oranges are good. Let \(\displaystyle \mathrm{G}_1,\mathrm{G}_2,\mathrm{G}_3\) be the events that the first, second and third orange drawn is good. Since the drawing is without replacement, apply the multiplication theorem for three events: \[\mathrm{P}(\mathrm{G}_1\cap \mathrm{G}_2\cap \mathrm{G}_3)=\mathrm{P}(\mathrm{G}_1)\,\mathrm{P}(\mathrm{G}_2\mid \mathrm{G}_1)\,\mathrm{P}(\mathrm{G}_3\mid \mathrm{G}_1\cap \mathrm{G}_2). \] There are $\displaystyle 15$ oranges, $\displaystyle 12$ of them good. After each good orange is removed, both the number of good oranges and the total fall by one: \[\mathrm{P}(\mathrm{G}_1)=\frac{12}{15},\qquad \mathrm{P}(\mathrm{G}_2\mid \mathrm{G}_1)=\frac{11}{14},\qquad \mathrm{P}(\mathrm{G}_3\mid \mathrm{G}_1\cap \mathrm{G}_2)=\frac{10}{13}. \] Therefore \[\mathrm{P}(\text{approved})=\frac{12}{15}\times\frac{11}{14}\times\frac{10}{13}=\frac{1320}{2730}=\frac{44}{91}. \] The probability that the box is approved for sale is \(\displaystyle \dfrac{44}{91}\).

The sample space for one toss of a coin together with one throw of a die is \[\mathrm{S}=\{(\mathrm{H},1),(\mathrm{H},2),\dots,(\mathrm{H},6),(\mathrm{T},1),\dots,(\mathrm{T},6)\},\qquad n(\mathrm{S})=2\times 6=12, \] all $\displaystyle 12$ outcomes equally likely.A = 'head on the coin' \(\displaystyle =\{(\mathrm{H},1),\dots,(\mathrm{H},6)\}\), so \(\displaystyle \mathrm{P}(\mathrm{A})=\dfrac{6}{12}=\dfrac{1}{2}\).B = '$\displaystyle 3$ on the die' \(\displaystyle =\{(\mathrm{H},3),(\mathrm{T},3)\}\), so \(\displaystyle \mathrm{P}(\mathrm{B})=\dfrac{2}{12}=\dfrac{1}{6}\).\(\displaystyle \mathrm{A}\cap\mathrm{B}=\{(\mathrm{H},3)\}\), so \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})=\dfrac{1}{12}\).Now apply the test for independence: \[\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}=\mathrm{P}(\mathrm{A}\cap\mathrm{B}). \] Since \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})=\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})\), the events A and B are independent.

The sample space is \(\displaystyle \mathrm{S}=\{1,2,3,4,5,6\}\) with all outcomes equally likely; $\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$ are red and $\displaystyle 4$, $\displaystyle 5$, $\displaystyle 6$ are green.A = 'the number is even' \(\displaystyle =\{2,4,6\}\), so \(\displaystyle \mathrm{P}(\mathrm{A})=\dfrac{3}{6}=\dfrac{1}{2}\).B = 'the number is red' \(\displaystyle =\{1,2,3\}\), so \(\displaystyle \mathrm{P}(\mathrm{B})=\dfrac{3}{6}=\dfrac{1}{2}\).\(\displaystyle \mathrm{A}\cap\mathrm{B}=\{2\}\) ($\displaystyle 2$ is the only number that is both even and red), so \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})=\dfrac{1}{6}\).Test for independence: \[\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\neq\frac{1}{6}=\mathrm{P}(\mathrm{A}\cap\mathrm{B}). \] Since \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})\neq \mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})\), the events A and B are not independent.

Use the test: E and F are independent if and only if \(\displaystyle \mathrm{P}(\mathrm{E}\cap\mathrm{F})=\mathrm{P}(\mathrm{E})\,\mathrm{P}(\mathrm{F})\).Compute the product of the given probabilities: \[\mathrm{P}(\mathrm{E})\,\mathrm{P}(\mathrm{F})=\frac{3}{5}\times\frac{3}{10}=\frac{9}{50}. \] Compare with the given intersection, written over the same denominator: \[\mathrm{P}(\mathrm{E}\cap\mathrm{F})=\frac{1}{5}=\frac{10}{50}. \] Since \(\displaystyle \dfrac{10}{50}\neq\dfrac{9}{50}\), we have \(\displaystyle \mathrm{P}(\mathrm{E}\cap\mathrm{F})\neq\mathrm{P}(\mathrm{E})\,\mathrm{P}(\mathrm{F})\).Hence E and F are not independent.

'Not A and not B' is the event \(\displaystyle \mathrm{A}'\cap\mathrm{B}'\). By De Morgan's law, \[\mathrm{A}'\cap\mathrm{B}'=(\mathrm{A}\cup\mathrm{B})', \] so \[\mathrm{P}(\mathrm{A}'\cap\mathrm{B}')=1-\mathrm{P}(\mathrm{A}\cup\mathrm{B}). \] First find \(\displaystyle \mathrm{P}(\mathrm{A}\cup\mathrm{B})\) by the addition theorem, using the given value of \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})\) (do not assume independence — here \(\displaystyle \mathrm{P}(\mathrm{A})\mathrm{P}(\mathrm{B})=\frac18=\mathrm{P}(\mathrm{A}\cap\mathrm{B})\), so they happen to be independent, but the given value is all we need): \[\mathrm{P}(\mathrm{A}\cup\mathrm{B})=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{2+4-1}{8}=\frac{5}{8}. \] Hence \[\mathrm{P}(\text{not A and not B})=1-\frac{5}{8}=\frac{3}{8}. \] \(\displaystyle \mathrm{P}(\mathrm{A}'\cap\mathrm{B}')=\dfrac{3}{8}\).

'Not A or not B' is the event \(\displaystyle \mathrm{A}'\cup\mathrm{B}'\). By De Morgan's law, \[\mathrm{A}'\cup\mathrm{B}'=(\mathrm{A}\cap\mathrm{B})', \] so the given information determines \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})\): \[\mathrm{P}(\mathrm{A}'\cup\mathrm{B}')=1-\mathrm{P}(\mathrm{A}\cap\mathrm{B})=\frac{1}{4}\ \Longrightarrow\ \mathrm{P}(\mathrm{A}\cap\mathrm{B})=1-\frac{1}{4}=\frac{3}{4}. \] This is the step students miss: the phrase 'not A or not B' must first be converted to the complement of \(\displaystyle \mathrm{A}\cap\mathrm{B}\).Now apply the test for independence: \[\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})=\frac{1}{2}\times\frac{7}{12}=\frac{7}{24},\qquad \mathrm{P}(\mathrm{A}\cap\mathrm{B})=\frac{3}{4}=\frac{18}{24}. \] Since \(\displaystyle \dfrac{18}{24}\neq\dfrac{7}{24}\), i.e. \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})\neq\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})\), the events A and B are not independent.(Worth noticing: \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})=\frac34>\frac12=\mathrm{P}(\mathrm{A})\), which no genuine pair of events can satisfy — the data are only illustrative, but the independence test still answers the question asked.)