# NCERT Solutions Class 12 Maths Chapter-13 (Probability)Exercise 13.1

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-13 (Probability)Exercise 13.1 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Class 12 Mathematics

Chapter-13 (Probability)

Questions and answers given in practice

Chapter-13 (Probability)

### Exercise 13.1

Q1. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P(F|E)

Answer. It is given that P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2

Q2. Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

Answer. It is given that P(B) = 0.5 and P(A ∩ B) = 0.32

Q3. If P(A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)

Answer. It is given that P(A) = 0.8, P(B) = 0.5, and P(B|A) = 0.4 (i) P (B|A) = 0.4  (iii) P(A∪B) = P(A) + P(B) – P(A∩B) ⇒ P(A∪B) = 0.8 + 0.5 – 0.32 = 0.98

Q4. Evaluate P(A ∪ B), if .

Q5. If  , find (i) P(A∩B) (ii) P(A|B) (iii) P(B|A) Determine P(E|F).

Answer. It is given that  (i) $\begin{array}{l}\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\frac{7}{11}\\ \therefore \mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{7}{11}\\ ⇒\frac{6}{11}+\frac{5}{11}-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{7}{11}\\ ⇒\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{11}{11}-\frac{7}{11}=\frac{4}{11}\end{array}$ (ii) It is known that, $P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$ $⇒\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\frac{4}{11}}{\frac{5}{11}}=\frac{4}{5}$ (iii) It is known that , $\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)}$

Q6. A coin is tossed three times, where (i) E : head on third toss , F : heads on first two tosses (ii) E : at least two heads , F : at most two heads (iii) E : at most two tails , F : at least one tail

Answer. (i)$\phantom{\rule{1em}{0ex}}E=\left\{HHH,HTH,THH,TH\right\}\phantom{\rule{0ex}{0ex}}F=\left\{HHH,HHT\right\}$$\mathrm{E}\cap \mathrm{F}=\left\{\mathrm{H}\mathrm{H}\mathrm{H}\right\}$  $\mathrm{E}=\left\{\mathrm{H}\mathrm{H}\mathrm{H},\mathrm{H}\mathrm{H}\mathrm{T},\mathrm{H}\mathrm{T}\mathrm{H},\mathrm{T}\mathrm{H}\mathrm{H}\right\}$ (ii) $\mathrm{F}=\left\{\mathrm{H}\mathrm{H}\mathrm{T},\mathrm{H}\mathrm{T}\mathrm{H},\mathrm{H}\mathrm{T}\mathrm{T},\mathrm{T}\mathrm{H}\mathrm{H},\mathrm{T}\mathrm{H}\mathrm{T},\mathrm{T}\mathrm{H},\mathrm{T}\mathrm{T}\mathrm{T}\right\}$ $\mathrm{E}\cap \mathrm{F}=\left\{\mathrm{H}\mathrm{H}\mathrm{T},\mathrm{H}\mathrm{T}\mathrm{H},\mathrm{T}\mathrm{H}\mathrm{H}\right\}$  $P\left(E|F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}=\frac{\frac{3}{8}}{\frac{7}{8}}=\frac{3}{7}$ (iii) $\mathrm{E}=\left\{\mathrm{H}\mathrm{H}\mathrm{H},\mathrm{H}\mathrm{H}\mathrm{T},\mathrm{H}\mathrm{T},\mathrm{H}\mathrm{T}\mathrm{H},\mathrm{T}\mathrm{H}\mathrm{H},\mathrm{T}\mathrm{H}\mathrm{T},\mathrm{T}\mathrm{H}\right\}$ $\left\{\mathrm{H}\mathrm{H}\mathrm{T},\mathrm{H}\mathrm{T}\mathrm{T},\mathrm{H}\mathrm{T}\mathrm{H},\mathrm{T}\mathrm{H}\mathrm{H},\mathrm{T}\mathrm{H}\mathrm{T},\mathrm{T}\mathrm{H},\mathrm{T}\mathrm{T}\mathrm{T}\right\}$  $\begin{array}{c}\mathrm{E}=\left\{\mathrm{H}\mathrm{H}\right\}\mathrm{F}=\left\{\mathrm{T}\mathrm{T}\right\}\\ \therefore \phantom{\rule{1em}{0ex}}\mathrm{E}\cap \mathrm{F}=\mathrm{\Phi }\end{array}$

Q7. Two coins are tossed once, where (i) E : tail appears on one coin, F : one coin shows head (ii) E : no tail appears, F : no head appears

Answer. If two coins are tossed once, then the sample space S is S = {HH, HT, TH, TT} (i) E = {HT, TH} F = {HT, TH} $\begin{array}{l}\therefore \mathrm{E}\cap \mathrm{F}=\left\{\mathrm{H}\mathrm{T},\mathrm{T}\mathrm{H}\right\}\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{2}{8}=\frac{1}{4}\\ \mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{2}{8}=\frac{1}{4}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}=\frac{2}{2}=1\end{array}$ (ii) E = {HH} F = {TT} ∴ E ∩ F = Φ P (F) = 1 and P (E ∩ F) = 0

Q8. A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses

Answer. If a die is thrown three times, then the number of elements in the sample space will be 6 × 6 × 6 = 216  $F=\left\{\left(6,5,1\right),\left(6,5,2\right),\left(6,5,3\right),\left(6,5,4\right),\left(6,5,5\right),\left(6,5,6\right)\right\}$ $\therefore \mathrm{E}\cap \mathrm{F}=\left\{\left(6,5,4\right)\right\}$

Q9. Mother, father and son line up at random for a family picture E : son on one end, F : father in middle .

Answer. If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be S = {MFS, MSF, FMS, FSM, SMF, SFM} E = {MFS, FMS, SMF, SFM} F = {MFS, SFM} $\mathrm{E}\cap \mathrm{F}=\left\{\mathrm{M}\mathrm{F}\mathrm{S},\mathrm{S}\mathrm{F}\mathrm{M}\right\}$

Q10. A black and a red dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Answer. Let the first observation be from the black die and second from the red die. When two dice (one black and another red) are rolled, the sample space S has 6 × 6 = 36 number of elements. 1. Let A: Obtaining a sum greater than 9  (a) The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B). $\therefore P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{\frac{2}{36}}{\frac{6}{35}}=\frac{2}{6}=\frac{1}{3}$ (b) E: Sum of the observations is 8. = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} F: Red die resulted in a number less than 4.$=\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(6,1\right),\left(6,2\right),\left(6,3\right)\end{array}\right\}$  The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E|F). Therefore,

Q11. A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5} Find (i) P(E|F) and P(F|E) (ii) P(E|G) and P(G|E) (iii) P((E ∪ F)|G) and P((E ∩ F)|G)

Answer. When a fair die is rolled, the sample space S will be S = {1, 2, 3, 4, 5, 6} It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5}