moochpuppy Posted March 31, 2004 Share Posted March 31, 2004 A die is rolled 10 times. Find the chances of -- a) getting 10 sixes 'B) not getting 10 sixes c) all the rolls showing 5 spots or less ______________________________________________ One ticket will be drawn at random from each of the two boxes shown below: (A) 1 2 3 ('B) 1 2 3 4 Find the chance that- a) The number drawn from A is larger than the one from B 'B) The number drawn from A equals the one from B c) The number drawn from A is smaller than the one from B ________________________________________________ Three cards are dealt from a well-shuffled deck. a) Find the chance that all of the cards are diamonds 'B) Find the chance that none of the cards are diamonds c) Find the chance that the cards are not all diamonds Any help please!!! Quote Link to comment Share on other sites More sharing options...
CrimsonWeltall Posted March 31, 2004 Share Posted March 31, 2004 A die is rolled 10 times. Find the chances of -- a) getting 10 sixes 'B) not getting 10 sixes c) all the rolls showing 5 spots or less ...ok. The odds of getting the first 6 are obviously 1/6. The odds of getting it the second time are still 1/6. So getting two sixes is (1/6)(1/6) = 1/36...and so on. Part b is a gimme as soon as you get part a. Part c is simple. One ticket will be drawn at random from each of the two boxes shown below: (A) 1 2 3 ('B) 1 2 3 4 Find the chance that- a) The number drawn from A is larger than the one from B 'B) The number drawn from A equals the one from B c) The number drawn from A is smaller than the one from B Just write out every possibility. A1 vs B1 A1 vs B2 ... A3 vs B3 Then determine how many of them fit the conditions. Three cards are dealt from a well-shuffled deck. a) Find the chance that all of the cards are diamonds 'B) Find the chance that none of the cards are diamonds c) Find the chance that the cards are not all diamonds This one is tougher, but you just have to know that diamonds are 13/52 cards. If one gets picked, then diamonds are 12/51 cards. If not, diamonds are 13/51 cards. Quote Link to comment Share on other sites More sharing options...
Gene Honda Civic Posted March 31, 2004 Share Posted March 31, 2004 I'll give it a shot --- My name is the Cheat so I'm copying off of the asian kid. A die is rolled 10 times. Find the chances of -- a) getting 10 sixes = (1/6)^10 'B) not getting 10 sixes = (5/6)^10 c) all the rolls showing 5 spots or less = (5/6)^10 -- I think ______________________________________________ One ticket will be drawn at random from each of the two boxes shown below: (A) 1 2 3 ('B) 1 2 3 4 Find the chance that- a) The number drawn from A is larger than the one from B 'B) The number drawn from A equals the one from B c) The number drawn from A is smaller than the one from B ________________________________________________ Three cards are dealt from a well-shuffled deck. a) Find the chance that all of the cards are diamonds = (1/4)*(12/51)*(11/50) 'B) Find the chance that none of the cards are diamonds = (39/52)*(38/51)*(37/50) c) Find the chance that the cards are not all diamonds = (3/4)*(38/51) The second group of questions has me stumped.. Quote Link to comment Share on other sites More sharing options...
CrimsonWeltall Posted March 31, 2004 Share Posted March 31, 2004 a) getting 10 sixes = (1/6)^10 'B) not getting 10 sixes = (5/6)^10 Cheat, the question was 'not getting 10 sixes', so getting 9 sixes is still good. The answer is 1 - part a. Quote Link to comment Share on other sites More sharing options...
Gene Honda Civic Posted March 31, 2004 Share Posted March 31, 2004 Cheat, the question was 'not getting 10 sixes', so getting 9 sixes is still good. The answer is 1 - part a. you're right... I was distracted by the hot, yet surprisingly unintelligent cheerleader next to me. generalizations> Quote Link to comment Share on other sites More sharing options...
moochpuppy Posted March 31, 2004 Author Share Posted March 31, 2004 ...ok. The odds of getting the first 6 are obviously 1/6. The odds of getting it the second time are still 1/6. So getting two sixes is (1/6)(1/6) = 1/36...and so on. Part b is a gimme as soon as you get part a. Part c is simple. So you are saying: a) (1/6)^10 'B) (1/6)^10 c) (5/6)^10 Quote Link to comment Share on other sites More sharing options...
CrimsonWeltall Posted March 31, 2004 Share Posted March 31, 2004 a) (1/6)^10 'B) (1/6)^10 c) (5/6)^10 No. Quote Link to comment Share on other sites More sharing options...
Gene Honda Civic Posted March 31, 2004 Share Posted March 31, 2004 So you are saying: a) (1/6)^10 'B) (1/6)^10 c) (5/6)^10 B ) = 1 - (1/6)^10 Quote Link to comment Share on other sites More sharing options...
moochpuppy Posted March 31, 2004 Author Share Posted March 31, 2004 B ) = 1 - (1/6)^10 And why is that? Quote Link to comment Share on other sites More sharing options...
moochpuppy Posted March 31, 2004 Author Share Posted March 31, 2004 Just write out every possibility. A1 vs B1 A1 vs B2 ... A3 vs B3 Then determine how many of them fit the conditions. a ) 1/4 b ) 1/4 c ) 1/2 Quote Link to comment Share on other sites More sharing options...
Gene Honda Civic Posted March 31, 2004 Share Posted March 31, 2004 part a is the probability that a 6 comes up 10 times in a row.... part b asks for the probability that a 6 doesn't come up 10 times in a row... It can come up 9 times however. 1 represents an assured event, i.e the likelyhood that a 2 headed coin lands on heads... or in the case of a die landing on any number. The only roll that wouldn't make it a sure thing would be 10 6's in a row. Since you already know the probability of that event(part a), and the probability of an assured event, you can subtract part a from 1 to obtain the answer for part b Quote Link to comment Share on other sites More sharing options...
moochpuppy Posted March 31, 2004 Author Share Posted March 31, 2004 part a is the probability that a 6 comes up 10 times in a row.... part b asks for the probability that a 6 doesn't come up 10 times in a row... It can come up 9 times however. 1 represents an assured event, i.e the likelyhood that a 2 headed coin lands on heads... or in the case of a die landing on any number. The only roll that wouldn't make it a sure thing would be 10 6's in a row. Since you already know the probability of that event(part a), and the probability of an assured event, you can subtract part a from 1 to obtain the answer for part b Got it! Thanks Cheat and Crimson for your help!!! BTW, does my answer for the second question look correct? Quote Link to comment Share on other sites More sharing options...
Gene Honda Civic Posted March 31, 2004 Share Posted March 31, 2004 BTW, does my answer for the second question look correct? EDIT>>>>>>>> i'm dumb.... You're smart.... You got em right Quote Link to comment Share on other sites More sharing options...
Texsox Posted March 31, 2004 Share Posted March 31, 2004 I'll give it a shot --- My name is the Cheat so I'm copying off of the asian kid. Quote Link to comment Share on other sites More sharing options...
SuperSteve Posted March 31, 2004 Share Posted March 31, 2004 A die is rolled 10 times. Find the chances of -- a) getting 10 sixes 'B) not getting 10 sixes c) all the rolls showing 5 spots or less ______________________________________________ One ticket will be drawn at random from each of the two boxes shown below: (A) 1 2 3 ('B) 1 2 3 4 Find the chance that- a) The number drawn from A is larger than the one from B 'B) The number drawn from A equals the one from B c) The number drawn from A is smaller than the one from B ________________________________________________ Three cards are dealt from a well-shuffled deck. a) Find the chance that all of the cards are diamonds 'B) Find the chance that none of the cards are diamonds c) Find the chance that the cards are not all diamonds Any help please!!! :finger Quote Link to comment Share on other sites More sharing options...
ChiSox_Sonix Posted April 1, 2004 Share Posted April 1, 2004 ah, i'm taking a Probability Theory II class right now (Stat major), i would have helped had i seen this. Quote Link to comment Share on other sites More sharing options...
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