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 | Probability |
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Can someone give detailed solution to this question???? Q)A shop sells 10 lamps out of which 3 are defective .Sachin buys 4 lamps Find the probability that atleast 2 of the lamps that he buys work? |
| Re: Probability |
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hi karan, ans wud be .3483 correct me if wrng |
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hi karan this is my approach to this problem first we can find the non occurence of the event and subtract it from 1 case 1 : all the 4 four he bought is good we can select 4 out of 10 in 10c4 ways and the good one can be selected in 7c4 ways 7C4/10C4 case 2: 3 good and 1 defective 3 can be selected from 7 in 7C3 ways and 1 defective can be selected in 3C1 7C3*3C1/10C4 from the above two cases we get 1/6 + 1/2 which is 0.666 this is the non occurence of the event atleast two defective and the occurence is 1-0.6666=0.3334 hope i am right |
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i think answer should be 203/210. we have to find the prob that at least 2 of the lamps that he buys work. lets find if he don't get 2 or more lamps working this means that he buys 3 defective n 1 working. this can be done by 7 ways ..... as selecting working lamp in 7c1 ways and 3 not working lamps 3c3.... so P(A') = (7c1*3c3)/10C4=7*1 =7/210 so P(A)=prob that atleast 2 of the lamps that he buys work. P(A)=1- P(A') =1-7/210 =203/210 |
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TO VIVEK 231, THE CASE U HAVE WROTE IS THE FAVOURABLE CASES OF BUYING ATLEAST 2 WORKING. I THINK U HAVE FORGOT TO ADD ONE MORE CASE OF BUYING 2 WORKING AND 2 NOT WORKING LAMPS. CASE 3: 2 GOOD AND 2 DEFECTIVE 2 WORKING CAN BE SELECTED FROM 7 IN 7C2 WAYS AND 2 DEFECTIVE CAN BE SELECTED IN 3C2 WAYS 7C2*3C2/10C4=3/10 SO OVERALL 1/6+1/2+3/10 =29/30 SO THIS IS THE TOTAL P OF BUYING ATLEAST 2 WORKING LAMPS. |
| Re: Probability |
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Hi Kiran, shop sells 10 lamps in that 3 are defective.. so, good------> 7 defective-->3 sachin buys 4 lamps so total possibilities are--> 10c4==210 in those 4 lamps, atleast 2 lamps will work atleast 2 means 2 or 3 or all hebuys work probability for 2 work and 2 defective--> 7c2 * 3c2==63 probability for 3 work and 1 defective--> 7c3 * 3c1==105 probability for all (4) will work----> 7c4==35 solution is== 63+105+35/210==>203/210==>0967 if is there any wrong please intimate me, thank you... |
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11 coins are tossed simultaneously..what is the probability of getting head in 9th n 10th coin.? |
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11 coins are tossed simultaneously..what is the probability of getting head in 9th n 10th coin.? |
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| 2^11-3/2^11. |
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prob of defective lamp=(3/10) so P of working lamps=(7/10) now p of getting 2 lamp that does not work=3c2/10c2=1/15 hence p of getting at leat 2 lamp that works=(1-1/15)=14/15 or p of getting 2 lamp that does not work=(3/10)*(2/9)=1-15 Ans:14/15. |
| Re: Probability |
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Can u please confirm if the answer is 1/4. I solved it considering that anything Head or Tail can come at any location except 9th and 10th position. These positions should only have Head . So total arrangements of the 11 coins in this case would be 2.2.2.2.2.2.2.2.1.1.2 = 2(pow 9) Total arrangements would be 2 ( pow 11) So probability = 2 ( pow 9 ) / 2 ( pow 11 ) = 1/4 Please confirm the answer. |
| Re: Probability |
| Ans: 2^9/2^11 is 1/4 |
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Slight modfication: the prob of selecting defect Xone=(3/10) & the good Xone=(7/10) there to get atleat 2 good=(7*7*3*3/10000)+(7*7*7*3/10000)+(7*7*7*7/10000)=0.3871. Ans: 0.3871 |
