{"id":17503,"date":"2021-01-29T17:50:44","date_gmt":"2021-01-29T12:20:44","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=17503"},"modified":"2021-01-29T17:50:44","modified_gmt":"2021-01-29T12:20:44","slug":"mcq-questions-for-class-11-maths-chapter-16","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/mcq-questions-for-class-11-maths-chapter-16\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 16 Probability with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 16 Probability with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Probability Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 16 Probability Objective Questions.<\/p>\n

Probability Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Probability Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Probability Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Probability Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nEvents A and B are independent if
\n(a) P (A \u2229 B) = P (A\/B) P (B)
\n(b) P (A \u2229 B) = P (B\/A) P (A)
\n(c) P (A \u2229 B) = P (A) + P (B)
\n(d) P (A \u2229 B) = P (A) \u00d7 P (B)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) P (A \u2229 B) = P (A) \u00d7 P (B)
\nEvents are said to be independent if the occurrence or non-occurrence of one event does not affect the probability of the occurrence or non-occurrence of the other.
\nNow, by the multiplication theorem,
\nP(A \u2229 B) = P(A) \u00d7 P(B\/A) ………… 1
\nSince A and B are independent events,
\nSo, P(B\/A) = P(B)
\nFrom equation 1, we get
\nP(A \u2229 B) = P(A) \u00d7 P(B)<\/p>\n<\/details>\n

\n

Question 2.
\nA single letter is selected at random from the word PROBABILITY. The probability that it is a vowel is
\n(a) 2\/11
\n(b) 3\/11
\n(c) 4\/11
\n(d) 5\/11<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 3\/11
\nThere are 11 letters in the word PROBABILITY out of which 1 can be selected in 11<\/sup>C1<\/sub> ways.
\nSo, exhaustive number of cases = 11
\nThere are 3 vowels i.e. A, I, O
\nSo, the favorable number of cases = 3
\nHence, the required probability = 3\/11<\/p>\n<\/details>\n

\n

Question 3.
\nA die is rolled, find the probability that an even prime number is obtained
\n(a) 1\/2
\n(b) 1\/3
\n(c) 1\/4
\n(d) 1\/6<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 1\/6
\nWhen a die is rolled, total number of outcomes = 6 (1, 2, 3, 4, 5, 6)
\nTotal even number = 3 (2, 4, 6)
\nNumber of even prime number = 1 (2)
\nSo, the probability that an even prime number is obtained = 1\/6<\/p>\n<\/details>\n

\n

Question 4.
\nWhen a coin is tossed 8 times getting a head is a success. Then the probability that at least 2 heads will occur is
\n(a) 247\/265
\n(b) 73\/256
\n(c) 247\/256
\n(d) 27\/256<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 247\/256
\nLet x be number a discrete random variable which denotes the number of heads obtained in n (in this question n = 8)
\nThe general form for probability of random variable x is
\nP(X = x) = nCx \u00d7 px \u00d7 qn-x
\nNow, in the question, we want at least two heads
\nNow, p =q = 1\/2
\nSo, P(X \u2265 2) = 8<\/sup>C2<\/sub> \u00d7 (1\/2)\u00b2 \u00d7 (1\/2)8-2<\/sup>
\n\u21d2 P(X \u2265 2) = 8<\/sup>C2<\/sub> \u00d7 (1\/2)\u00b2 \u00d7 (1\/2)6<\/sup>
\n\u21d2 1 – P(X < 2) = 8<\/sup>C0<\/sub> \u00d7 (1\/2)0<\/sup> \u00d7 (1\/2)8<\/sup> + 8<\/sup>C1<\/sub> \u00d7 (1\/2)1<\/sup> \u00d7 (1\/2)8-1<\/sup>
\n\u21d2 1 – P(X < 2) = (1\/2)8<\/sup> + 8 \u00d7 (1\/2)1<\/sup> \u00d7 (1\/2)7<\/sup>
\n\u21d2 1 – P(X < 2) = 1\/256 + 8 \u00d7 (1\/2)8<\/sup>
\n\u21d2 1 – P(X < 2) = 1\/256 + 8\/256
\n\u21d2 1 – P(X < 2) = 9\/256
\n\u21d2 P(X < 2) = 1 – 9\/256
\n\u21d2 P(X < 2) = (256 – 9)\/256
\n\u21d2 P(X < 2) = 247\/256<\/p>\n<\/details>\n

\n

Question 5.
\nThe probability that the leap year will have 53 sundays and 53 monday is
\n(a) 2\/3
\n(b) 1\/2
\n(c) 2\/7
\n(d) 1\/7<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 1\/7
\nIn a leap year, total number of days = 366 days.
\nIn 366 days, there are 52 weeks and 2 days.
\nNow two days may be
\n(i) Sunday and Monday
\n(ii) Monday and Tuesday
\n(iii) Tuesday and Wednesday
\n(iv) Wednesday and Thursday
\n(v) Thursday and Friday
\n(vi) Friday and Saturday
\n(vii) Saturday and Sunday
\nNow in total 7 possibilities, Sunday and Monday both come together is 1 time.
\nSo probabilities of 53 Sunday and Monday in a leap year = 1\/7<\/p>\n<\/details>\n

\n

Question 6.
\nLet A and B are two mutually exclusive events and if P(A) = 0.5 and P(B \u0305) =0.6 then P(AUB) is
\n(a) 0
\n(b) 1
\n(c) 0.6
\n(d) 0.9<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 0.9
\nGiven, A and B are two mutually exclusive events.
\nSo, P(A \u2229 B) = 0
\nAgain given P(A) = 0.5 and P(B \u0305) = 0.6
\nP(B) = 1 – P(B \u0305) = 1 – 0.6 = 0.4
\nNow, P(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n\u21d2 P(A \u222a B) = P(A) + P(B)
\n\u21d2 P(A \u222a B) = 0.5 + 0.4 = 0.9<\/p>\n<\/details>\n

\n

Question 7.
\nSeven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals
\n(a) 1\/2
\n(b) 7\/15
\n(c) 2\/15
\n(d) 1\/3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 7\/15
\nWhile placing 7 while balls in a row, total gaps = 8
\n3 black balls can be placed in 8 gaps = C = (8 \u00d7 7 \u00d7 6)\/(3 \u00d7 2 \u00d7 1) = 8 \u00d7 7 = 56
\nSo, the total number of ways of arranging white and black balls such that no two black balls are adjacent = 56 \u00d7 3! \u00d7 7!
\nActual number of arrangement possible with 7 white and 3 black balls = (7 + 3)! = 10!
\nSo, the required Probability = (56 \u00d7 3! \u00d7 7!)\/10!
\n= (56 \u00d7 3! \u00d7 7!)\/(10 \u00d7 9 \u00d7 8 \u00d7 7!)
\n= (56 \u00d7 3!)\/(10 \u00d7 9 \u00d7 8)
\n= (56 \u00d7 3 \u00d7 2 \u00d7 1)\/(10 \u00d7 9 \u00d7 8)
\n= (7 \u00d7 3 \u00d7 2 \u00d7 1)\/(10 \u00d7 9)
\n= (7 \u00d7 2)\/(10 \u00d7 3)
\n= 7\/(5 \u00d7 3)
\n= 7\/15<\/p>\n<\/details>\n

\n

Question 8.
\nThe events A, B, C are mutually exclusive events such that P (A) = (3x + 1)\/3, P (B) = (x – 1)\/4 and P (C) = (1 – 2x)\/4. The set of possible values of x are in the interval
\n(a) [1\/3, 1\/2]
\n(b) [1\/3, 2\/3]
\n(c) [1\/3, 13\/3]
\n(d) [0, 1]<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) [1\/3, 1\/2]
\nP(A) = (3x + 1)\/3
\nP(B) = (x – 1)\/4
\nP(C) = (1 – 2x)\/4
\nThese are mutually exclusive events.
\n\u21d2 -1 \u2264 3x \u2264 2, -3 \u2264 x \u2264 1, -1 \u2264 2x \u2264 1
\n\u21d2 -1\/3 \u2264 x \u2264 2\/3, -2 \u2264 x \u2264 1, -1\/2 \u2264 x \u2264 1\/2
\nAlso, 0 \u2264 (3x + 1)\/3 + (x – 1)\/4 + (1 – 2x)\/4 \u2264 1
\n\u21d2 1\/3 \u2264 x \u2264 13\/3
\n\u21d2 max {-1\/3, -3, -1\/2, 1\/3} \u2264 x \u2264 min {2\/3, 1\/2, 1, 13\/3}
\n\u21d2 1\/3 \u2264 x \u2264 1\/2
\n\u21d2 x \u2208 [1\/3, 1\/2]<\/p>\n<\/details>\n

\n

Question 9.
\nA bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. The probability that none of the balls drawn is blue is
\n(a) 10\/21
\n(b) 11\/21
\n(c) 2\/7
\n(d) 5\/7<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 10\/21
\nTotal number of balls = 2 + 3 + 2 = 7
\nTwo balls are drawn.
\nNow, P(none of them is blue) = 5<\/sup>C2<\/sub> \/ 7<\/sup>C2<\/sub>
\n= {(5 \u00d7 4)\/(2 \u00d7 1)}\/{(7 \u00d7 6)\/(2 \u00d7 1)}
\n= (5 \u00d7 4)\/(7 \u00d7 6)
\n= (5 \u00d7 2)\/(7 \u00d7 3)
\n= 10\/21<\/p>\n<\/details>\n

\n

Question 10.
\nIf 4-digit numbers greater than 5000 are randomly formed from the digits 0, 1, 3, 5 and 7, then the probability of forming a number divisible by 5 when the digits are repeated is
\n(a) 1\/5
\n(b) 2\/5
\n(c) 3\/5
\n(d) 4\/5<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 2\/5
\nGiven digits are 0, 1, 3, 5, 7
\nNow we have to form 4 digit numbers greater than 5000.
\nSo leftmost digit is either 5 or 7.
\nWhen digits are repeated
\nNumber of ways for filling left most digit = 2
\nNow remaining 3 digits can be filled = 5 \u00d7 5 \u00d7 5
\nSo total number of ways of 4 digits greater than 5000 = 2 \u00d7 5 \u00d7 5 \u00d7 5 = 250
\nAgain a number is divisible by 5 if the unit digit is either 0 or 5. So there are 2 ways to fill the unit place.
\nSo total number of ways of 4 digits greater than 5000 and divisible by 5 = 2 \u00d7 5 \u00d7 5 \u00d7 2 = 100
\nNow probability of 4 digit numbers greater than 5000 and divisible by 5
\n= 100\/250
\n= 2\/5<\/p>\n<\/details>\n

\n

Question 11.
\nEvents A and B are said to be mutually exclusive iff
\n(a) P (A U B) = P (A) + P (B)
\n(b) P (A \u2229 B) = P (A) \u00d7 P (B)
\n(c) P(A U B) = 0
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) P (A U B) = P (A) + P (B)
\nIf A and B are mutually exclusive events,
\nThen P(A \u2229 B) = 0
\nNow, by the addition theorem,
\nP(A U B) = P(A) + P(B) – P(A \u2229 B)
\n\u21d2 P(A U B) = P(A) + P(B)<\/p>\n<\/details>\n

\n

Question 12.
\nTwo numbers are chosen from {1, 2, 3, 4, 5, 6} one after another without replacement. Find the probability that the smaller of the two is less than 4.
\n(a) 4\/5
\n(b) 1\/15
\n(c) 1\/5
\n(d) 14\/15<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 4\/5
\nTotal number of ways of choosing two numbers out of six = 6<\/sup>C2<\/sub> = (6 \u00d7 5)\/2 = 3 \u00d7 5 = 15
\nIf smaller number is chosen as 3 then greater has choice are 4, 5, 6
\nSo, total choices = 3
\nIf smaller number is chosen as 2 then greater has choice are 3, 4, 5, 6
\nSo, total choices = 4
\nIf smaller number is chosen as 1 then greater has choice are 2, 3, 4, 5, 6
\nSo, total choices = 5
\nTotal favourable case = 3 + 4 + 5 = 12
\nNow, required probability = 12\/15 = 4\/5<\/p>\n<\/details>\n

\n

Question 13.
\nIf the integers m and n are chosen at random between 1 and 100, then the probability that the number of the from 7m<\/sup> + 7\u207f is divisible by 5 equals
\n(a) 1\/4
\n(b) 1\/7
\n(c) 1\/8
\n(d) 1\/49<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 1\/4
\nSince m and n are selected between 1 and 100,
\nHence total sample space = 100 \u00d7 100
\nAgain, 71 = 7, 72 = 49, 73 = 343, 74 = 2401, 75 = 16807, etc
\nHence 1, 3, 7 and 9 will be the last digit in the power of 7.
\nNow, favourable number of case are
\n\u2192 1,1 1,2 1,3 …………. 1,100
\n2,1 2,2 2,3 …………. 2,100
\n3,1 3,2 3,3 …………. 3,100
\n……………….
\n………………
\n100,1 100,2 100,3 …………. 100,100
\nNow, for m = 1, n = 3, 7, 11, ………, 97
\nSo, favourable cases = 25
\nAgain for m = 2, n = 4, 8, 12, ………, 100
\nSo, favourable cases = 25
\nHence for every m, favourable cases = 25
\nSo, total favourable cases = 100 \u00d7 25
\nRequired Probability = (100 \u00d7 25)\/(100 \u00d7 100)
\n= 25\/100
\n= 1\/4<\/p>\n<\/details>\n

\n

Question 14.
\nA card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn. Then the probability that they both are diamonds is
\n(a) 84\/452
\n(b) 48\/452
\n(c) 84\/425
\n(d) 48\/425<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 48\/425
\nTotal number of cards = 52 and one card is lost.
\nCase 1: if lost card is a diamond card
\nTotal number of cards = 51
\nNumber of diamond cards = 12
\nNow two cards are drawn.
\nP(both cards are diamonds) = 12<\/sup>C2<\/sub> \/ 51<\/sup>C2<\/sub>
\nTotal number of cards = 52 and one card is lost.
\nCase 2: If lost card is not a diamond card
\nTotal number of cards = 51
\nNumber of diamond cards = 13
\nNow two cards are drawn.
\nP(both cards are diamonds) = 13<\/sup>C2<\/sub> \/ 51<\/sup>C2<\/sub>
\nNow probability that both cards are diamond = 12<\/sup>C2<\/sub> \/ 51<\/sup>C2<\/sub> + 13<\/sup>C2<\/sub> \/ 51<\/sup>C2<\/sub>
\n= (12<\/sup>C2<\/sub> + 13<\/sup>C2<\/sub>) \/ 51<\/sup>C2<\/sub>
\n= {(12 \u00d7 11)\/(2 \u00d7 1) + (13 \u00d7 12)\/(2 \u00d7 1)}\/{(51 \u00d7 50)\/(2 \u00d7 1)}
\n= (12 \u00d7 11 + 13 \u00d7 12)\/(51 \u00d7 50)
\n= (132 + 156)\/2550
\n= 288\/2550
\n= 96\/850 (288 and 2550 divided by 3)
\n= 48\/425 (96 and 850 divided by 2)
\nSo probability that both cards are diamond is 48\/425<\/p>\n<\/details>\n

\n

Question 15.
\nThe probability that when a hand of 7 cards is drawn from a well-shuffled deck of 52 cards, it contains 3 Kings is
\n(a) 1\/221
\n(b) 5\/716
\n(c) 9\/1547
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 9\/1547
\nTotal number of cards = 52
\nNumber of king card = 4
\nNow, 7 cards are drawn from 52 cards.
\nP (3 cards are king) = {4<\/sup>C3<\/sub> \u00d7 48<\/sup>C4<\/sub>}\/ 52<\/sup>C7<\/sub>
\n= {4 \u00d7 (48 \u00d7 47 \u00d7 46 \u00d7 45)\/(4 \u00d7 3 \u00d7 2 \u00d7 1)}\/{(52 \u00d7 51 \u00d7 50 \u00d7 49 \u00d7 48 \u00d7 47 \u00d7 46)\/(7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1)}
\n= {4 \u00d7 (48 \u00d7 47 \u00d7 46 \u00d7 45) \u00d7 (7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1)}\/{(4 \u00d7 3 \u00d7 2 \u00d7 1) \u00d7 {(52 \u00d7 51 \u00d7 50 \u00d7 49 \u00d7 48 \u00d7 47 \u00d7 46)}
\n= (7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 45)\/(52 \u00d7 51 \u00d7 50 \u00d7 49)
\n= (6 \u00d7 5 \u00d7 4 \u00d7 45)\/(52 \u00d7 51 \u00d7 50 \u00d7 7)
\n= (6 \u00d7 4 \u00d7 45)\/(7 \u00d7 52 \u00d7 51 \u00d7 10)
\n= (6 \u00d7 45)\/(7 \u00d7 13 \u00d7 51 \u00d7 10)
\n= (6 \u00d7 3)\/(7 \u00d7 13 \u00d7 17 \u00d7 2)
\n= (3 \u00d7 3)\/(7 \u00d7 13 \u00d7 17)
\n= 9\/1547<\/p>\n<\/details>\n

\n

Question 16.
\nA die is rolled, then the probability that an even number is obtained is
\n(a) 1\/2
\n(b) 2\/3
\n(c) 1\/4
\n(d) 3\/4<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 1\/2
\nWhen a die is rolled, total number of outcomes = 6 (1, 2, 3, 4, 5, 6)
\nTotal even number = 3 (2, 4, 6)
\nSo, the probability that an even number is obtained = 3\/6 = 1\/2<\/p>\n<\/details>\n

\n

Question 17.
\nSix boys and six girls sit in a row at random. The probability that the boys and girls sit alternatively is
\n(a) 1\/462
\n(b) 11\/462
\n(c) 5\/121
\n(d) 7\/123<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 1\/462
\nGiven, 6 boys and 6 girls sit in a row at random.
\nThen, the total number of arrangement of 6 boys and 6 girls = arrangement of 12 persons = 12!
\nNow, boys and girls sit alternatively.
\nSo, the total number of arrangement = 2 \u00d7 6! \u00d7 6!
\nNow, P(boys and girls sit alternatively) = (2 \u00d7 6! \u00d7 6!)\/12!
\n= (2\u00d7 6 \u00d7 5! \u00d7 6!)\/(12 \u00d7 11!)
\n= (5! \u00d7 6!)\/11!
\n= (5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 \u00d7 6!)\/(11 \u00d7 10 \u00d7 9 \u00d7 8 \u00d7 7 \u00d7 6!)
\n= (5 \u00d7 4 \u00d7 3 \u00d7 2)\/(11 \u00d7 10 \u00d7 9 \u00d7 8 \u00d7 7)
\n= (4 \u00d7 3)\/(11 \u00d7 9 \u00d7 8 \u00d7 7)
\n= 3\/(11 \u00d7 9 \u00d7 2 \u00d7 7)
\n= 1\/(11 \u00d7 3 \u00d7 2 \u00d7 7)
\n= 1\/462<\/p>\n<\/details>\n

\n

Question 18.
\nTwo dice are thrown the events A, B, C are as follows A: Getting an odd number on the first die. B: Getting a total of 7 on the two dice. C: Getting a total of greater than or equal to 8 on the two dice. Then AUB is equal to
\n(a) 15
\n(b) 17
\n(c) 19
\n(d) 21<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 21
\nWhen two dice are thrown, then total outcome = 6 \u00d7 6 = 36
\nA: Getting an odd number on the first die.
\nA = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4),(3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4),(5, 5), (5, 6)}
\nTotal outcome = 18
\nB: Getting a total of 7 on the two dice.
\nB = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
\nTotal outcome = 6
\nC: Getting a total of greater than or equal to 8 on the two dice.
\nC = {(2, 6), (3, 5), (3, 6), (4, 4),(4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6),(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
\nTotal outcome = 15
\nNow n(A U B) = n(A) + n(B) – n(A \u2229 B)
\n\u21d2 n(A U B) = 18 + 6 – 3
\n\u21d2 n(A U B) = 21<\/p>\n<\/details>\n

\n

Question 19.
\nLet A and B are two mutually exclusive events and if P(A) = 0.5 and P(B \u0305) =0.6 then P(AUB) is
\n(a) 0
\n(b) 1
\n(c) 0.6
\n(d) 0.9<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 0.9
\nGiven, A and B are two mutually exclusive events.
\nSo, P(A \u2229 B) = 0
\nAgain given P(A) = 0.5 and P(B \u0305) = 0.6
\nP(B) = 1 – P(B \u0305) = 1 – 0.6 = 0.4
\nNow, P(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n\u21d2 P(A \u222a B) = P(A) + P(B)
\n\u21d2 P(A \u222a B) = 0.5 + 0.4 = 0.9<\/p>\n<\/details>\n

\n

Question 20.
\nA certain company sells tractors which fail at a rate of 1 out of 1000. If 500 tractors are purchased from this company, what is the probability of 2 of them failing within first year
\n(a) e-1\/2<\/sup> \/2
\n(b) e-1\/2<\/sup> \/4
\n(c) e-1\/2<\/sup> \/8
\n(d) none of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) e-1\/2<\/sup> \/8
\nThis question is based on Poisson distribution.
\nNow, \u03bb = np = 500\u00d7(1\/1000) = 500\/1000 = 1\/2
\nNow, P(x = 2) = {e-1\/2<\/sup> \u00d7 (1\/2)\u00b2}\/2! = e-1\/2<\/sup> \/(4 \u00d7 2) = e-1\/2<\/sup> \/8<\/p>\n<\/details>\n

\n

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 16 Probability with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the …<\/p>\n

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