| BACK PROBLEMS: 1)An biased die is tossed.Find the probability of getting a multiple of 3? Sol: Here we have sample space S={1,2,3,4,5,6}. Let E be the event of getting a multiple of 3. Then E={3,6}. P(E) =n(E)/n(S). n(E) =2, n(S) =6. P(E) =2/6 P(E) =1/3. 2)In a simultaneous throw of a pair of dice,find the probability of getting a total more than 7? Sol: Here we have sample space n(S) =6*6 =36. Let E be the event of getting a total more than 7. ={(1,6),(2,5),(3,4),(4,3)(5,2),(6,1)(2,6),(3,5),(4,4),(5,3),(6,2),(4,5),(5,4), (5,5),(4,6),(6,4)} n(E) =15 P(E) = n(E)/n(S) = 15/36. P(E) = 5/12. 3)A bag contains 6 white and 4 black balls .Two balls are drawn at random .Find the probability that they are of the same colour? Sol: Let S be the sample space. Number of ways for drawing two balls out of 6 white and 4 red balls = 10C2 =10!/(8!*2!) = 45. n(S) =45. Let E =event of getting both balls of the same colour. Then n(E) =number of ways of drawing ( 2balls out of 6) or (2 balls out of 4). = 6C2 +4C2 = 6!/(4!*2!) + 4!/(2! *2!) = 6*5/2 +4 *3/2 =15+6 =21. P(E) =n(E)/n(S) =21/45 =7/45. 4)Two dice are thrown together .What is the probability that the sum of the number on the two faces is divisible by 4 or 6? Sol: n(S) = 6*6 =36. E be the event for getting the sum of the number on the two faces is divisible by 4 or 6. E={(1,3)(1,5)(2,4?)(2,2)(3,5)(3,3)(2,6)(3,1)(4,2)(4,4)(5,1)(5,3)(6,2)(6,6)} n(E) =14. Hence P(E) =n(E)/n(S) = 14/36. P(E) = 7/18 5)Two cards are drawn at random from a pack of 52 cards What is the probability that either both are black or both are queens? Sol: total number of ways for choosing 2 cards from 52 cards is =52C2 =52 !/(50!*2!) = 1326. Let A= event of getting bothe black cards. Let B= event of getting bothe queens AnB=Event of getting queens of black cards n(A) =26C2. We have 26 black cards from that we have to choose 2 cards. n(A) =26C2=26!/(24!*2!) = 26*25/2=325 from 52 cards we have 4 queens. n(B) = 4C2 = 4!/(2!* 2!) =6 n(AnB) =2C2. =1 P(A) = n(A) /n(S) =325/1326 P(B) = n(B)/n(S) = 6/1326 P(A n B) = n(A n B)/n(S) = 1/1326 P(A u B) = P(A) +P(B) -P(AnB) = 325/1326 + 6/1326 -1/1326 = 330/1326 P(AuB) = 55/221 6)Two diced are tossed the probability that the total score is a prime number? Number of total ways n(S) =6 * 6 =36 E =event that the sum is a prime number. Then E={(1,1)(1,2)(1,4)(1,6)(2,1)(2,3)(2,5)(3,2)(3,4)(4,1)(4,3)(5,2)(5,6)(6,1)(6,5)} n(E) =15 P(E) =n(E)/n(S) = 15/36 P(E) = 5/12 7)Two dice are thrown simultaneously .what is the probability of getting two numbers whose product is even? Sol : In a simultaneous throw of two dice ,we have n(S) = 6*6 = 36 E =Event of getting two numbers whose product is even E ={(1,2)(1,4)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,2)(3,4)(3,6)(4,1) (4,2)(4,3)(4,4)(4,5)(4,6)(5,2)(5,4)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)} n(E) = 27 P(E) = n(E)/n(S) = 27 /36 P(E) =3/4 probability of getting two numbers whose product is even is equals to 3/4. 8)In a lottery ,there are 10 prozes and 25 blanks.A lottery is drawn at random. what is the probability of getting a prize ? Sol: By drawing lottery at random ,we have n(S) =10C1+25C1 = 10+25 = 35. E =event of getting a prize. n(E) =10C1 =10 out of 10 prozes we have to get into one prize .The number of ways 10C1. n(E) =10 n(S) =35 P(E) =n(E)/n(S) =10/35 = 2/7 Probability is 2/7. 9)In a class ,30 % of the students offered English,20 % offered Hindi and 10 %offered Both.If a student is offered at random ,wha5t is the probability that he has offered English or Hindi? English offered students =30 %. Hindi offered students =20% Both offered students =10 % Then only english offered students E =30 -10 =20 % only Hindi offered students S =20 -10 % = 10 % All the students =100% =E +S +E or S 100 =20 +10 + E or S +E and S Hindi or English offered students =100 -20-10-10 =60 % Probability that he has offered English or Hindi =60/100 = 2/5 10) A box contains 20 electricbulbs ,out of which 4 are defective ,two bulbs are chosen at random from this box.What is the probability that at least one of these is defective ? Sol: out of 20 bulbs ,4 bulbs are defective. 16 bulbs are favourable bulbs. E = event for getting no bulb is defective. n(E) =16 C 2 out of 16 bulbs we have to choose 2 bulbs randomly .so the number of ways =16 C 2 n(E) =16 C2 n(S) =20 C 2 P(E) =16 C2/20C2 = 12/19 probability of at least one is defective + probability of one is non defective =1 P(E) + P(E) =1 12/19 +P(E) =1 P(E’) =7/19 11)A box contains 10 block and 10 white balls.What is the probability of drawing two balls of the same colour? Sol: Total number of balls =10 +10 =20 balls Let S be the sample space. n(S) =number of ways drawing 2 balls out of 20 = 20 C2 = 20 !/(18! *2!) = 190. Let E =event of drawing 2 balls of the same colour. n(E) =10C2+ 10C2 = 2(10 C2) = 90 P(E) =n(E)/n(S) P(E) =90/190 = 9/19 12) A bag contains 4 white balls ,5 red and 6 blue balls .Three balls are drawn at random from the bag.What is the probability that all of them are red ? Sol: Let S be the sample space. Then n(S) =number of ways drawing 3 balls out of 15. =15 C3. =455 Let E =event of getting all the 3 red balls. n(E) = 5 C3 =5C2 = 10 P(E) =n(E) /n(S) =10/455 =2/91. 13)From a pack of 52 cards,one card is drawn at random.What is the probability that the card is a 10 or a spade? Sol: Total no of cards are 52. These are 13 spades including tne and there are 3 more tens. n(E) =13+3 = 16 P(E) =n(E)/n(S). =16/52 P(E) =4/13. 14) A man and his wife appear in an interview for two vacancies in the same post. The probability of husband's selection is 1/7 and the probabililty of wife's selection is 1/5. What is the probabililty that only one of them is selected? Sol: let A =event that the husband is selected. B = event that the wife is selected. E = Event for only one of them is selected. P(A) =1/7 and p(B) =1/5. P(A') =Probability of husband is not selected is =1-1/7=6/7 P(B') =Probaility of wife is not selected =1-1/5=4/7 P(E) =P[(A and B') or (B and A')] = P(A and B') +P(B and A') = P(A)P(B') + P(B)P(A') = 1/7*4/5 + 1/5 *6/7 P(E) =4/35 +6/35=10/35 =2/7 15)one card is drawn at random from a pack of 52 cards.What is the probability that the card drawn is a face card? Sol: There are 52 cards,out of which there 16 face cards. P(getting a face card) =16/52 = 4/13 16) The probability that a card drawn from a pack of 52 cards will be a diamond or a king? Sol: In 52 cards 13 cards are diamond including one king there are 3 more kings. E event of getting a diamond or a king. n(E) =13 +3 = 16 P(E) =n(E) /n(S) =16/52 =4/13 17) Two cards are drawn together from apack of 52 cards.What is the probability that one is a spade and one is a heart ? Sol: S be the sample space the n (S) =52C2 =52*51/2 =1326 let E =event of getting 2 kings out of 4 kings n(E) =4C2 = 6 P(E) =n(E)/n(S) =6/1326 =1/221 18) Two cards are drawn together from a pack of 52 cards.What is the probability that one is a spade and one is a heart? Sol: Let S be the sample space then n(S) =52C2 =1326 E = Event of getting 1 spade and 1 heart. n(E) =number of ways of choosing 1 spade out of 13 and 1 heart out of 13. = 13C1*13C1 =169 P(E)= n(E)/n(S) =169/1326 =13/102. 19) Two cards are drawn from a pack of 52 cards .What is the probability that either both are Red or both are Kings? Sol: S be the sample space. n(S) =The number of ways for drawing 2 cards from 52 cards. n(S) =52C2 =1326 E1 be the event of getting bothe red cards. E2 be the event of getting both are kings. E1nE2 =Event of getting 2 kings of red cards. We have 26 red balls.From 26 balls we have to choose 2 balls. n(E1) =26C2 = 26*25/2 =325 We have 4 kings .out of 4 kings,we have to choosed 2 balls. n(E2) =4C2 =6 n(E1nE2) =2C2 =1 P(E1) = n(E1)/n(S) =325/1326 P(E2) =n(E2)/n(S) =6/1326 P(E1nE2) =n(E1nE2)/n(S) =1/1326 P(both red or both kings) = P(E1UE2) = P(E1) +P(E2)-P(E1nE2) =325/1326 +6/1326 -1/1326 =330/1326 =55/221 BACK |