Alternative English Year - 2017 Full Marks : 100
Class 12 Ncert Solutions Probability Maths
Exercise 13.1
1. 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).
Sol: We have P(E|F) =\(\frac{P(E∩F)}{P(F)}\) = \(\frac{0.2}{0.3}=\frac{2}{3}\)
and P(F|E) = \(\frac{P(E∩F)}{P(E)}\) = \(\frac{0.2}{0.6}=\frac{2}{6}=\frac{1}{3}\)
2. Compute P(A|B), if P(B) = 0.5 and P(A∩B) = 0.32
Sol: We have P(A|B) = \(\frac{P(A∩B)}{P(B)}\) = \(\frac{0.32}{0.5}=\frac{32}{50}=\frac{16}{25}\)
3. 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)
Sol: (i) P(A∩B)
P(B|A) = \(\frac{ P(A∩B) }{P(A)}\)
\(\implies\) P(A∩B) = P(B|A)P(A)=0.4 \(\times\)0.8= 0.32
(ii) P(A|B) = \(\frac{P(A∩B)}{P(B)}\) =\(\frac{0.32}{0.5}=\frac{32}{50}=0.64\)
(iii) P(A∪B) = P(A) + P(B) - P(A∩B) = 0.8 + 0.5 - 0.32 = 0.98
4. Evaluate P(A∪B) , if 2P(A) = P(B) = \(\frac{5}{13}\) and P(A|B) = \(\frac{2}{5}\)
Sol: P(A|B) = \(\frac{P(A∩B)}{P(B)}\)
\(\implies\) P(A∩B) = P(A|B)P(B) = \(\frac{2}{5}\times\frac{5}{13}=\frac{10}{65}\)
\(\therefore\) P(A∪B) = P(A) + P(B) - P(A∩B) = \(\frac{5}{26}+\frac{5}{13}-\frac{10}{65}\)
= \(\frac{25+50-20}{130}\)
= \(\frac{55}{130}\)
= \(\frac{11}{26}\)
5. If P(A) =\(\frac{6}{11}\), P(B) =\(\frac{5}{11}\) and P(A∪B)=\(\frac{7}{11}\), find
(i) P(A∩B) (ii) P(A|B) (iii) P(B|A)
Sol: (i) We have P(A∪B) = P(A) + P(B) - P(A∩B)
\(\implies\)P(A∩B) = P(A) + P(B)- P(A∪B) = \(\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}\)
(ii) P(A|B) = \(\frac{P(A∩B)}{P(B)}\) = \({\frac{4}{11}\over\frac{5}{11}}=\frac{4}{5}\)
(iii) P(B|A) = \(\frac{P(A∩B)}{P(A)}\) = \({\frac{4}{11}\over\frac{6}{11}}=\frac{4}{6}=\frac{2}{3}\)
Determine P(E|F) in exercise 6 to 9.
6. 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
Sol : A coin is tossed three times. So the sample space of this event can be given as
{HHH, HHT, HTT, HTH, TTT, TTH, THT, THH}
(i) Sample space of event F is = {HHH,HHT}
Sample space of event E ={HHT}
n(E∩F ) = 1
n(F) = 2
Probabilty of getting E(i.e head on the third toss i.e HHH) considering the sample space F = \(\frac{1}{2}\)
\(\therefore\) Probability of E given that F has occurred i.e P(E|F)=\(\frac{1}{2}\)
(ii) E: at least two heads. F : at most two heads
Sample space of F ={HHH,HHT,HTT,HTH,THT,THH,TTH}
Sample space of E ={HHH,HHT,THH}
n(E∩F ) = 3
n(F) = 7
Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = \(\frac{3}{7}\)
(iii) E: at most two tails, F : at least one tail
Sample space of F = {TTT, TTH, THT, HTT, HHT, HTH, THH}
Sample space of E ={ TTH, THT, HTT, HHT, HTH, THH}
n(E∩F ) = 6
n(F) = 7
Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = \(\frac{6}{7}\)
7. 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
Sol: Sample space of a two coin tossed once = {HH, HT, TH, TT}
(i) E: tail appears on one coin, F : one coin shows head
Sample space of F={HT,TH}
Sample space of E = {HT,TH}
Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = \(\frac{P(E∩F)}{P(F)}=\frac{2}{2}=1\)
(ii) E : no tail appears, F: no head appears
Sample space of F = {TT}
Sample space of E= {HH}
Here, P(E∩F) =\(\frac{0}{4}\) since there is no common element.
P(F) = \(\frac{1}{4}\)
Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = \(\frac{P(E∩F)}{P(F)}=\frac{0}{1}=0\)
8. A die is thrown three times,
E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses
Sol: Sample space of F = {651, 652, 653, 654, 655, 656}
Sample space of E = {114,124,...164,214,224,...,264,314,...,364,414...464,514...564,614...664}
Here, E∩F =1 since only one common element
P(E∩F) = \(\frac{1}{216}\)
P(F) = \(\frac{6}{216}\)
Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = \(\frac{P(E∩F)}{P(F)}=\frac{1}{6}\)
Class 12 Ncert Solutions Probability Maths
E : son on one end, F: father in middle
Sol: Sample space of the event of lining up at random for a famliy picture= {SFM, SMF, FSM, FMS, MSF, MFS}
Sample space of F = {MFS,SFM}
Sample Space of E ={SFM, SMF, MFS,FMS}
Here, E∩F =2 since only two common element
P(E∩F) =\(\frac{2}{6}\)
P(F) = \(\frac{2}{6}\)
Probability of getting E considering the sample space F orProbability of E given that F has occurred i.e. P(E|F) = \(\frac{P(E∩F)}{P(F)}={\frac{2}{6}\over\frac{2}{6}}=1\)
10. 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 thet the red die resulted in anumber less than 4.
Sol: (a) Let E be the event of obtaining a sum greater than 9
F be the event that the black die resulted in a 5.
Sample space of F ={51,52,53,54,55,56}
Sample space of E = {55,56} because 5+5=10 and 5+6 = 11 are the only ones greater than 9
E∩F = 2
P(E∩F) = \(\frac{2}{36}\)
P(F) = \(\frac{6}{36}\)
The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5
or Probability of E given that F has occurred i.e. P(E|F) = \(\frac{P(E∩F)}{P(F)}={\frac{2}{36}\over\frac{6}{36}}=\frac{1}{3}\)
(b) Let E be the event of obtaining a sum 8
F be the event that the black die resulted in a number less than 4.
Sample space of F ={11,12,13,21,22,23,31,32,33,41,42,43,51,52,53,61,62,63}
Sample space of E = {53,62} because 5+5=10 and 6+5 = 11 are the only ones greater than 9
E∩F=2
P(E∩F) = \(\frac{2}{36}\)
P(F) = \(\frac{18}{36}\)
The conditional probability of obtaining a sum 8, given that the red die resulted in a number less than 4
or Probability of E given that F has occurred i.e. P(E|F) = \(\frac{P(E∩F)}{P(F)}={\frac{2}{36}\over\frac{18}{36}}=\frac{1}{9}\)
11. 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)
Sol: Given E = {1,3,5}, F={2,3} and G = {2,3,4,5}
\(\therefore E\cap F\) = {3} i.e. n(E\(\cap F\)) = 1
\(\therefore E\cap G\) = {3,5} i.e. n(E\(\cap F\)) = 2
(i) P(E|F) = \(\frac{n(E\cap F)}{n(F)}=\frac{1}{2}\)
P(F|E) = \(\frac{n(E\cap F)}{n(E)}=\frac{1}{3}\)
(ii) P(E|G) = \(\frac{n(E\cap G)}{n(G)}=\frac{2}{4}= \frac{1}{2}\)
P(G|E) = \(\frac{n(E\cap G)}{n(E)}=\frac{2}{3}= \frac{2}{3}\)
(iii) P((E∩F)|G) = \(\frac{n(E\cap F\cap G)}{n(G)}=\frac{1}{4}= \frac{1}{4}\)
P(F|G)= \(\frac{n(F\cap G)}{n(G)}=\frac{2}{4}= \frac{1}{2}\)
P((E∪F)|G) = P(E|G) + P(F|G) - P((E∩F)|G) = \(\frac{1}{2}+\frac{1}{2} -\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)
12. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that(i) the youngest is a girl, (ii) at least one is girl?
Sol: Let b stand for boy and g stand for girl. The sample space of the experiment is
S = {(b,b),(g,b),(b,g),(g,g)}
Let E be the event that both the children are girls.
(i) Let F be the event that the youngest children is a girl.
Then E = {(b,b)} and F = { (g,g),(b,g)} \(\therefore\)E∩F = 1
Now, P(E∩F) = \(\frac{1}{4}\)
Therefore, P(E|F) = \(\frac{P(E∩F)}{P(F)}={\frac{1}{4}\over\frac{2}{4}}=\frac{1}{2}\)
(ii) Let F be the event that at least one children is a girl
Then E = {(b,b)} and F = { (g,g),(b,g),(g,b)}
Now, P(E∩F) = \(\frac{1}{4}\)
Therefore, P(E|F) = \(\frac{P(E∩F)}{P(F)}={\frac{1}{4}\over\frac{3}{4}}=\frac{1}{3}\)
13. An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/ False questions, 500 easy multiple choice questions and 400 difficlut multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Sol: Total no. of multiple choice question = 500 easy + 400 difficult = 900
Total no. of easy multiple choice question = 500
The probability that the selected random question will be easy given that it is a multiple choice question is = \(\frac{500}{900}=\frac{5}{9}\)
14. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ' the sum of numbers on the dice is 4'.
Sol: Let E be the event that the sum of numbers on the dice is 4 and
F be the event that the two numbers appearing on throwing two dice are different.
Sample space of throwing two dice consist of \(6^2\) = 36 elements
Sample space of F = {12,13,14,15,16,21,23,24,25,26,31,32,34,35,36,41,42,43,45,46,51,52,53,54,56,61,62,63,64,65}
Sample space of E = {13,31}
E∩F = 2
Now, P(E∩F) = \(\frac{2}{36}\)
P(E|F) =\({\frac{2}{36}\over\frac{30}{36}}=\frac{1}{15}\)
15. Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given thet ' at least one die shows a 3'.
Sol:
Let F be an event that a multiple of 3 comes up while throwing a die and E be the event that a coin shows tail when tossed.
F = {31,32...36,61,62...66}
E = {1H,1T,2H,2T,4H,4T,5H,5T}
n(E∩F) = 0
n(F) = 12
Probability of E given that F has occurered i.e. P(E|F) = \(\frac{n(E∩F)}{n(F) }=\frac{0}{12}=0\)
Choose the correct answer :
16. If P(A) =\(\frac{1}{2}\), P(B) = 0, then P(A|B) is
(A) 0 (B) \(\frac{1}{2}\) (C) not defined (D) 1
Sol: P(A|B) = \(\frac{P(A\cap B)}{P(B)}= \frac{P(A\cap B)}{0}\) = not defined
The correct option is (C) not defined
17. If A and B are events such that P(A|B) = P(B|A) then
(A) A\(\subset\)B but A\(\neq\) B (B) A = B (C) A\(\cap\) B = \(\phi\) (D) P(A) = P(B)
Sol : P(A|B) = \(\frac{P(A\cap B)}{P(B)}\)
P(B|A) = \(\frac{P(A\cap B)}{P(A)}\)
\(\therefore\) \(\frac{P(A\cap B)}{P(B)}\) = \(\frac{P(A\cap B)}{P(A)}\)
\(\implies\) P(A)= P(B)
The correct answer is option (D).
Class 12 Ncert Solutions Probability Maths
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Previous exercises
Exercise 13.1
Exercise 13.5
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