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Well, area of smaller circle = (πr²)/4 and area between larger circle and smaller circle is (3πr²)/4. Total area is πr². So the areas clearly show the probabilities of finding a dot . . .
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Oh yeah! Thanks Zishi for taking the trouble to explain me that.Well, area of smaller circle = (πr²)/4 and area between larger circle and smaller circle is (3πr²)/4. Total area is πr². So the areas clearly show the probabilities of finding a dot . . .
Hey even i have a soubt in this can anyone helpHi friends, i've got this question. I've been struggling hard with it all night but didnt reach the answer.
Could someone please help me out.
Five men, 3 women and 2 children were queuing in a row at a bus stop. Calculate the number of ways they can queue up if only 2 of the women are to stand next to each other.
Ans : 1693440
Thanks in advance to the intellectuals who gonna help me .
Hey even i have a soubt in this can anyone help
smzimran
For question four part 2,is it that you don't understand the question,or how to solve the answer.Can anybody please answer my doubts?
View attachment 14749
I can't solve rather.For question four part 2,is it that you don't understand the question,or how to solve the answer.
Haha okay,here goesI can't solve rather.
Nope, didn't get confused, your explanations are crystal clear to me. Thanks a lot mate.Haha okay,here goes
So basically when they say a seven day period,it means a week.So,the question asks if four weeks are randomly chosen from how many ever wnter weeks there are,what is the probability that exactly three of the four chosen weeks have had at least one day with more than 20 cm of snow(That is three of four weeks have had 1,2,3,4,5,6 or 7 days of more than 20cm snow)i.e all the 7 days should have had 20cm or more of snow in all three weeks.
Going back to the question,first Find the probability that a week will have "at least 1 day with more than 20cm of snow falling"
For this you can do 1-(probability 0 days have more than 20cm snow) = 1-(7C0 * (0.79)^7) =0.808
Then you have that n=4
p=0.808
q=0.102
Then you can just use the binomial distribution method
Exactly 3 weeks would mean (4C3 * (0.808^3) * (0.192))
and then you get 0.405
Hope this didn't confuse you
Your welcomeNope, didn't get confused, your explanations are crystal clear to me. Thanks a lot mate.
Yup. I'm doing 3 AS+A2 and 2 AS at one go. wbu?Your welcome
Are you doing your AS level for the first time this Nov
I'm doing 3AS,actually resitsYup. I'm doing 3 AS+A2 and 2 AS at one go. wbu?
Hey do not be discouraged, you know the french saying : "reculer pour mieux sauter" which means "back to jump higher".I'm doing 3AS,actually resits
The may june results were shocking.
I have a question.. Please someone help me.
1. Ivan throws three fair dice.
(i) List all the possible scores on the three dice which give a total score of 5, and hence show that
1
[3]
the probability of Ivan obtaining a total score of 5 is 36 .
(ii) Find the probability of Ivan obtaining a total score of 7.
Hey thanks. But its not enough clear in my mindProbability of getting 5's (3,1,1),(1,1,3),(1,3,1),(2,1,2),(2,2,1),(1,2,2) so we get 6{(1/6)*(1/6)*(1/6)} which is 1/36
For 7 we have, (5,1,1), (1,5,1), (1,1,5), (4,2,1), (2,4,1), (1,2,4), (2,1,4), (4,1,2), (1,4,2), (3,3,1), (1,3,1), (1,1,3), (2,2,3), (3,2,2), (2,3,2) which is 15{(1/6)*(1/6)*(1/6)}=5/72
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