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21.(iii)    In a class of $\small 60$ students, $\small 30$ opted for NCC, $\small 32$ opted for NSS and $\small 24$ opted for both NCC and NSS. If one of these students is selected at random, find the probability that

(iii) The student has opted NSS but not NCC.

Let A be the event that student opted for NCC and B be the event that the student opted for NSS. Given,  n(S) = 60, n(A) = 30, n(B) =32, n(A  B) = 24 Therefore, P(A) =  P(B) =  P(A  B) =  (iii) Now, Probability that the student has opted NSS but not NCC = P(B  A' ) = P(B-A) We know, P(B-A) = P(B) -  P(A  B)  Hence,the probability that the student has opted NSS but not NCC is

21.(ii)    In a class of $\small 60$ students, $\small 30$ opted for NCC, $\small 32$ opted for NSS and $\small 24$ opted for both NCC and NSS. If one of these students is selected at random, find the probability that

(ii) The student has opted neither NCC nor NSS.

Let A be the event that student opted for NCC and B be the event that the student opted for NSS. Given,  n(S) = 60, n(A) = 30, n(B) =32, n(A  B) = 24 Therefore, P(A) =  P(B) =  P(A  B) =  (ii) Now, Probability that the student has opted neither NCC nor NSS = P(A'  B' ) We know, P(A'  B' ) = 1 - P(A  B) [De morgan's law] And, P(A  B) = P(A)+ P(B) - P(A  B)     P(A'  B' )  Hence,the probability...

21.(i)   In a class of  $\small 60$  students, $\small 30$ opted for NCC, $\small 32$ opted for NSS and  $\small 24$ opted for both NCC and NSS. If one of these students is selected at random, find the probability that

(i) The student opted for NCC or NSS.

Let A be the event that student opted for NCC and B be the event that the student opted for NSS. Given,  n(S) = 60, n(A) = 30, n(B) =32, n(A  B) = 24 Therefore, P(A) =  P(B) =  P(A  B) =  (i) We know, P(A  B) = P(A)+ P(B) - P(A  B)    Hence,the probability that the student opted for NCC or NSS is

20. The probability that a student will pass the final examination in both English and Hindi is  $\small 0.5$ and the probability of passing neither is $\small 0.1$. If the probability of passing the English examination is  $\small 0.75$, what is the probability of passing the Hindi examination?

Let A be the event that the student passes English examination and B be the event that the students pass Hindi examination. Given,  P(A)=0.75, P(A   B) = 0.5, P(A'  B') =0.1 We know, P(A'  B') = 1 - P(A  B)  P(A  B) = 1 - 0.1 = 0.9 Also, P(A  B) = P(A)+ P(B) - P(A  B)  P(B) = 0.9 - 0.75 + 0.5 = 0.65 Hence,the probability of passing the Hindi examination is 0.65

19.    In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is $\small 0.8$ and the probability of passing  the second examination is  $\small 0.7$ . The probability of passing atleast one of them is  $\small 0.95$ . What is the probability of passing both?

Let A be the event that the student passes the first examination and B be the event that the students passes the second examination. P(A  B) is probability of passing at least one of the examination. Therefore,  P(A  B) = 0.95 , P(A)=0.8, P(B)=0.7 We know, P(A  B) = P(A)+ P(B) - P(A   B)  P(A  B) = 0.8 + 0.7 - 0.95 = 1.5 -0.95 = 0.55 Hence,the probability that the student will pass both the...

18.    In Class XI of a school  $\small 40\%$ of the students study Mathematics and $\small 30\%$  study Biology. $\small 10\%$ of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Let M denote the event that the student is studying Mathematics and B denote the event that the student is studying Biology And total students in the class be 100. Given, n(M) = 40  P(M) =  n(B) = 30 P(M) =  n(M  B) = 10 P(M) =  We know, P(A  B) = P(A)+ P(B) - P(A  B)  P(M  B) = 0.4 + 0.3 - 0.1 = 0.6 Hence, the probability that he will be studying Mathematics or Biology is 0.6

17(iii) A and B are events such that  P(A) $\small =0.42$ , P(B) $\small =0.48$  and P(A and B) $\small =0.16$. Determine (iii)  P(A or B)

Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16 (iii) We know,     = 0.74

17.(ii)  A and B are events such that P(A) $\small =0.42$,  P(B) $\small =0.48$ and  P(A and B) $\small =0.16$. Determine (ii) P(not B)

Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16 (ii)   Therefore, P(not B) = 0.52

17.(i)   A and B are events such that P(A) $\small =0.42$,  P(B) $\small =0.48$ and  P(A and B) . $\small =0.16$Determine (i)  P(not A)

Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16 (i)   Therefore, P(not A) = 0.58

16.    Events E and F are such that P(not E or not F) $\small =0.25$, State whether E and F are mutually exclusive.

Given,  For A and B to be mutually exclusive,  Now,  We know,  Hence, E and F are not mutually exclusive.

15.(ii)   If E and F are events such that  $P(E)=\frac{1}{4}$  ,  $P(F)=\frac{1}{2}$  and   $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$ find  (ii) P(not E and not F).

Given, ,      and     To find : We know, And     Therefore,

15.(i)   If E and F are events such that   $P(E)=\frac{1}{4}$,   $P(F)=\frac{1}{2}$   and    $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$,  find (i)  P(E or F)

Given, ,      and     To find : We know,        Therefore,

14.  Given  $P(A)=\frac{3}{5}$   and    $P(B)=\frac{1}{5}$.  Find  $P(A\hspace{1mm}or\hspace{1mm}B)$, if  $A$ and  $B$ are mutually exclusive  events.

Given,     and  To find : We know,    [Since A and B are mutually exclusive events.]   Therefore,

13    Fill in the blanks in following table:

 $P(A)$ $P(B)$ $P(A\cap B)$ $P(A\cup B)$ (i) $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{15}$ $...$ (ii) $0.35$ $...$ $0.25$ $0.6$ (iii) $0.5$ $0.35$ $...$ $0.7$

We know,    (i)  =   =  (ii)     (iii)       (i) (ii) 0.5 (iii) 0.15

12.(ii)     Check whether the following probabilities  $P(A)$  and  $P(B)$  are consistently defined

(ii)   $P(A)=0.5,P(B)=0.4,P(A\cup B)=0.8$

(ii)   Given,  We know,   P(A  B) = P(A)+ P(B) - P(A  B) 0.8 = 0.5 + 0.4 - P(A  B)  P(A  B) = 0.9 - 0.8 = 0.1 Therefore, P(A  B) < P(A) and P(A  B) < P(B) , which satisfies the condition. Hence, the probabilities are consistently defined

12.(i)   Check whether the following probabilities  $P(A)$  and  $P(B)$ are consistently defined:

(i)   $P(A)=0.5,P(B)=0.7.P(A\cap B)=0.6$

(i) Given, Now P(A  B) > P(A)   (Since A  B is a subset of A, P(A  B) cannot be more than P(A))   Therefore, the given probabilities are not consistently defined.

10.(ii)   A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (ii) a consonant

Given, ‘ASSASSINATION’ No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1 No. of letters = 13 No. of consonants = {4 S's,2 N's,T} = 7 One letter is selected: n(S) =  = 13 Let E be the event of getting a consonant. n(E) =  = 7

10.(i)  A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (i) a vowel

Given, ‘ASSASSINATION’ No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1 No. of letters = 13 No. of vowels = {3 A's,2 I's,O} = 6 One letter is selected: n(S) =  = 13 Let E be the event of getting a vowel. n(E) =  = 6

9.   If  $\small \frac{2}{11}$  is the probability of an event, what is the probability of the event ‘not A’.

Given, P(E) =  We know,  P(not E) = P(E') = 1 - P(E)   =   =

8.(ix)   Three coins are tossed once. Find the probability of getting

(ix) atmost two tails

Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8] Let E be the event of getting atmost 2 tails = Event of getting 2 or less tails = {HHT, HTH, THH, TTH, HTT, THT}  n(E) = 6      The required probability of getting atmost 2 tails is .
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