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<html><table height="500" width="1000" border="2"><TR height="5" width="1000"><strong><center><marquee><font color="Green"><h1>APTITUDE</h1></center></strong></font></TR></marquee><TR><TD align="left" width="200" valign="top"><table><TR><a href="numbers.html"><strong>Numbers</strong></a></TR><br><TR><a href="hcf.html"><strong>H.C.F and L.C.M</strong></a></TR><br><TR><a href="dec.html" ><strong>Decimal Fractions</strong></a></TR><br><TR><a href="simplification.html"><strong>Simplification</strong></a></TR><br><TR><a href="squareandcuberoot.html" ><strong>Square and Cube roots</strong></a></TR><br><TR><a href="average.html" ><strong>Average</strong></a></TR><br><TR><a href="pnumbers.html" ><strong>Problems on Numbers</strong></a></TR><br><TR><a href="problemsonages.html"><strong>Problems on Ages</strong></a></TR><br><TR><a href="surdsandindices.html"><strong>Surds and Indices</strong></a></TR><br><TR><a href="percent.html" ><strong>Percentage</strong></a></TR><br><TR><a href="profitandloss.html" ><strong>Profit and Loss</strong></a></TR><br><TR><a href="ratioandproportion.html" ><strong>Ratio And Proportions</strong></a></TR><br><TR><a href="partnership.html"><strong>Partnership</strong></a></TR><br><TR><a href="chainrule1.html"><strong>Chain Rule</strong></a></TR><br><TR><a href="timeandwork.html" ><strong>Time and Work</strong></a></TR><br><TR><a href="pipesandcisterns.html" ><strong>Pipes and Cisterns</strong></a></TR><br><TR><a href="timeanddistance.html"><strong>Time and Distance</strong></a></TR><br><TR><a href="trains.html" ><strong>Trains</strong></a></TR><br><TR><a href="boats.html"><strong>Boats and Streams</strong></a></TR><br><TR><a href="alligation.html"><strong>Alligation or Mixture </strong></a></TR><br><TR><a href="simple.html" ><strong>Simple Interest</strong></a></TR><br><TR><a href="CI.html"><strong>Compound Interest</strong></a></TR><br><TR><a href=""><strong>Logorithms</strong></a></TR><br><TR><a href="areas.html" ><strong>Areas</strong></a></TR><br><TR><a href="volume.html"><strong>Volume and Surface area</strong></a></TR><br><TR><a href="races.html" ><strong>Races and Games of Skill</strong></a></TR><br><TR><a href="calendar.html" ><strong>Calendar</strong></a></TR><br><TR><a href="clocks.html" ><strong>Clocks</strong></a></TR><br><TR><a href="" ><strong>Stocks ans Shares</strong></a></TR><br><TR><a href="true.html" ><strong>True Discount</strong></a></TR><br><TR><a href="banker1.html" ><strong>Bankers Discount</strong></a></TR><br><TR><a href="oddseries.html"><strong>Oddmanout and Series</strong></a></TR><br><TR><a href=""><strong>Data Interpretation</strong></a></TR><br><TR><a href="probability.html"><strong>probability</strong></a></TR><br><TR><a href="percom1.html" ><strong>Permutations and Combinations</strong></a></TR><br><TR><a href="pinkivijji_puzzles.html" ><strong>Puzzles</strong></a></TR></table></TD><TD ><a href="probability.html"><b>BACK</b></a><br><br><font size="5"><b><center>PROBABILITY</center></b></font><br><br><br><br> <font size="4"><b>PROBLEMS:</b><br><br>1)An biased die is tossed.Find the probability of getting a multiple of 3?<br>Sol: Here we have sample space S={1,2,3,4,5,6}.<br> Let E be the event of getting a multiple of 3.<br> Then E={3,6}.<br> P(E) =n(E)/n(S).<br> n(E) =2,<br> n(S) =6.<br> P(E) =2/6<br> P(E) =1/3.<br><br>2)In a simultaneous throw of a pair of dice,find the probability of getting a total<br> more than 7?<br>Sol: Here we have sample space n(S) =6*6 =36.<br>Let E be the event of getting a total more than 7.<br> ={(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),<br>(5,5),(4,6),(6,4)}<br> n(E) =15<br> P(E) = n(E)/n(S)<br> = 15/36.<br> P(E) = 5/12.<br><br>3)A bag contains 6 white and 4 black balls .Two balls are drawn at random .Find the<br> probability that they are of the same colour?<br>Sol: Let S be the sample space.<br> Number of ways for drawing two balls out of 6 white and 4 red balls = 10C2<br> <br> =10!/(8!*2!)<br> <br> = 45.<br> n(S) =45.<br> Let E =event of getting both balls of the same colour.<br> Then<br> n(E) =number of ways of drawing ( 2balls out of 6) or (2 balls out of 4).<br> = 6C2 +4C2<br> = 6!/(4!*2!) + 4!/(2! *2!)<br> = 6*5/2 +4 *3/2<br> =15+6 =21.<br> P(E) =n(E)/n(S) =21/45 =7/45.<br><br>4)Two dice are thrown together .What is the probability that the sum of the number on<br> the two faces is divisible by 4 or 6?<br>Sol: n(S) = 6*6 =36.<br>E be the event for getting the sum of the number on the two faces is divisible by 4<br> or 6.<br>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)}<br>n(E) =14.<br>Hence P(E) =n(E)/n(S)<br> = 14/36.<br>P(E) = 7/18<br><br>5)Two cards are drawn at random from a pack of 52 cards What is the probability that<br> 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!)<br> <br> = 1326.<br>Let A= event of getting bothe black cards.<br>Let B= event of getting bothe queens<br> AnB=Event of getting queens of black cards<br> n(A) =26C2.<br> We have 26 black cards from that we have to choose 2 cards.<br> n(A) =26C2=26!/(24!*2!)<br> = 26*25/2=325<br> from 52 cards we have 4 queens.<br><br> n(B) = 4C2<br> = 4!/(2!* 2!) =6<br> n(AnB) =2C2. =1<br> P(A) = n(A) /n(S) =325/1326<br> P(B) = n(B)/n(S) = 6/1326<br> P(A n B) = n(A n B)/n(S) = 1/1326<br> P(A u B) = P(A) +P(B) -P(AnB)<br> = 325/1326 + 6/1326 -1/1326<br> = 330/1326<br> P(AuB) = 55/221<br><br>6)Two diced are tossed the probability that the total score is a prime number?<br>Number of total ways n(S) =6 * 6 =36<br>E =event that the sum is a prime number.<br>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)}<br>n(E) =15<br>P(E) =n(E)/n(S)<br> = 15/36<br> P(E) = 5/12<br><br>7)Two dice are thrown simultaneously .what is the probability of getting two numbers<br> whose product is even?<br>Sol : In a simultaneous throw of two dice ,we have n(S) = 6*6<br> = 36<br> E =Event of getting two numbers whose product is even<br> 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)<br>(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)}<br>n(E) = 27<br> P(E) = n(E)/n(S)<br> = 27 /36<br>P(E) =3/4<br>probability of getting two numbers whose product is even is equals to 3/4.<br><br><br>8)In a lottery ,there are 10 prozes and 25 blanks.A lottery is drawn at random.<br>what is the probability of getting a prize ?<br> Sol: By drawing lottery at random ,we have n(S) =10C1+25C1<br> = 10+25<br> = 35.<br>E =event of getting a prize.<br> n(E) =10C1 =10<br>out of 10 prozes we have to get into one prize .The number of ways 10C1.<br> n(E) =10<br> n(S) =35<br> P(E) =n(E)/n(S)<br> =10/35<br> = 2/7<br> Probability is 2/7.<br><br>9)In a class ,30 % of the students offered English,20 % offered Hindi and 10 %offered<br> Both.If a student is offered at random ,wha5t is the probability that <br>he has offered English or Hindi?<br> English offered students =30 %.<br> Hindi offered students =20%<br> Both offered students =10 %<br>Then only english offered students E =30 -10<br> =20 %<br>only Hindi offered students S =20 -10 %<br> = 10 %<br>All the students =100% =E +S +E or S<br> 100 =20 +10 + E or S +E and S<br>Hindi or English offered students =100 -20-10-10<br> =60 %<br>Probability that he has offered English or Hindi =60/100 = 2/5<br><br>10) A box contains 20 electricbulbs ,out of which 4 are defective ,two bulbs are <br>chosen at random from this box.What is the probability that at least <br>one of these is defective ?<br> Sol: out of 20 bulbs ,4 bulbs are defective.<br> 16 bulbs are favourable bulbs.<br> E = event for getting no bulb is defective.<br> n(E) =16 C 2<br>out of 16 bulbs we have to choose 2 bulbs randomly .so the number of ways =16 C 2<br> n(E) =16 C2<br> n(S) =20 C 2<br> P(E) =16 C2/20C2<br> = 12/19<br>probability of at least one is defective + probability of one is non defective =1<br> P(E) + P(E) =1<br> 12/19 +P(E) =1<br> P(E’) =7/19<br><br>11)A box contains 10 block and 10 white balls.What is the probability of drawing two<br> balls of the same colour?<br>Sol: Total number of balls =10 +10<br> =20 balls<br>Let S be the sample space.<br> n(S) =number of ways drawing 2 balls out of 20<br> = 20 C2<br> = 20 !/(18! *2!)<br> = 190.<br>Let E =event of drawing 2 balls of the same colour.<br> n(E) =10C2+ 10C2<br> = 2(10 C2)<br> = 90<br> P(E) =n(E)/n(S)<br>P(E) =90/190<br> = 9/19<br><br>12) A bag contains 4 white balls ,5 red and 6 blue balls .Three balls are drawn at<br> random from the bag.What is the probability that all of them are red ?<br><br>Sol: Let S be the sample space.<br> Then n(S) =number of ways drawing 3 balls out of 15.<br> =15 C3.<br> =455<br>Let E =event of getting all the 3 red balls.<br> n(E) = 5 C3 =5C2 <br> = 10<br>P(E) =n(E) /n(S) =10/455 =2/91.<br><br><br>13)From a pack of 52 cards,one card is drawn at random.What is the probability that<br> the card is a 10 or a spade?<br>Sol: Total no of cards are 52.<br> These are 13 spades including tne and there are 3 more tens.<br> n(E) =13+3 <br> = 16<br> P(E) =n(E)/n(S).<br> =16/52<br>P(E) =4/13.<br><br>14) A man and his wife appear in an interview for two vacancies in the same post.<br>The probability of husband's selection is 1/7 and the probabililty of wife's <br>selection is 1/5.<br>What is the probabililty that only one of them is selected?<br> Sol: let A =event that the husband is selected.<br> B = event that the wife is selected.<br> E = Event for only one of them is selected.<br> P(A) =1/7<br> and<br> p(B) =1/5.<br> P(A') =Probability of husband is not selected is =1-1/7=6/7<br> P(B') =Probaility of wife is not selected =1-1/5=4/7<br> P(E) =P[(A and B') or (B and A')]<br> = P(A and B') +P(B and A')<br> = P(A)P(B') + P(B)P(A')<br> = 1/7*4/5 + 1/5 *6/7<br> P(E) =4/35 +6/35=10/35 =2/7<br><br>15)one card is drawn at random from a pack of 52 cards.What is the probability<br> that the card drawn is a face card?<br><br>Sol: There are 52 cards,out of which there 16 face cards.<br> P(getting a face card) =16/52<br> = 4/13<br>16) The probability that a card drawn from a pack of 52 cards will be a diamond <br>or a king?<br><br> Sol: In 52 cards 13 cards are diamond including one king there are 3 more kings.<br> E event of getting a diamond or a king.<br> n(E) =13 +3<br> = 16<br>P(E) =n(E) /n(S) =16/52<br> =4/13<br><br>17) Two cards are drawn together from apack of 52 cards.What is the probability that<br> one is a spade and one is a heart ?<br>Sol: S be the sample space the n (S) =52C2 =52*51/2<br> =1326 <br> let E =event of getting 2 kings out of 4 kings <br> n(E) =4C2<br> = 6<br> P(E) =n(E)/n(S)<br> =6/1326<br> =1/221<br><br>18) Two cards are drawn together from a pack of 52 cards.What is the probability <br>that one is a spade and one is a heart?<br> Sol: Let S be the sample space then<br> n(S) =52C2<br> =1326<br> E = Event of getting 1 spade and 1 heart.<br> n(E) =number of ways of choosing 1 spade out of 13 and 1 heart out of 13.<br> = 13C1*13C1 =169<br>P(E)= n(E)/n(S)<br> =169/1326 =13/102.<br>19) Two cards are drawn from a pack of 52 cards .What is the probability that either<br> both are Red or both are Kings?<br>Sol: S be the sample space.<br> n(S) =The number of ways for drawing 2 cards from 52 cards.<br> n(S) =52C2<br> =1326<br> E1 be the event of getting bothe red cards.<br> E2 be the event of getting both are kings.<br> E1nE2 =Event of getting 2 kings of red cards.<br>We have 26 red balls.From 26 balls we have to choose 2 balls.<br> n(E1) =26C2<br> = 26*25/2<br> =325<br>We have 4 kings .out of 4 kings,we have to choosed 2 balls.<br> n(E2) =4C2<br> =6<br> n(E1nE2) =2C2 =1<br>P(E1) = n(E1)/n(S)<br> =325/1326<br>P(E2) =n(E2)/n(S)<br> =6/1326<br>P(E1nE2) =n(E1nE2)/n(S) =1/1326<br>P(both red or both kings) = P(E1UE2)<br> = P(E1) +P(E2)-P(E1nE2)<br> =325/1326 +6/1326 -1/1326<br> =330/1326 =55/221<br><br></font><a href="probability.html"><b>BACK</b></a></TD></TR></table></html>?? 快捷鍵說明
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