March 31, 200422 yr 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!!!
March 31, 200422 yr 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.
March 31, 200422 yr 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..
March 31, 200422 yr 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.
March 31, 200422 yr 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>
March 31, 200422 yr Author ...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
March 31, 200422 yr Author 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
March 31, 200422 yr 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
March 31, 200422 yr Author 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?
March 31, 200422 yr BTW, does my answer for the second question look correct? EDIT>>>>>>>> i'm dumb.... You're smart.... You got em right
March 31, 200422 yr I'll give it a shot --- My name is the Cheat so I'm copying off of the asian kid.
March 31, 200422 yr 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
April 1, 200422 yr ah, i'm taking a Probability Theory II class right now (Stat major), i would have helped had i seen this.
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.