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A level Statistics doubt??Post your doubts here!

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tag me in the sol as well

Part iv : You have P _ _ So thats 8C2 * 3! Because the three cards can take different orders = 168

Part v : You have P G _ Or _ P G Or G P _ Or _ G P So thats 7C 1 * 4
However they asked for do not have blabla so thats your answer in Part iii (9C3 * 3!) - (7*2*2) = 476
 
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A star Alice123
(v)really got to me
The pink and the green card is selected, so only one card is to be chosen from 7 remaining cards. But the condition is no pink and green next to each other-
There are your possible arrangements for the pink and green card being together.
P G X
G P X

X G P
X P G


So to choose 1 cards from 7 cards= 7C1= 7. there are four possible arrangements so 7*4= 28

So 28 arrangements from the 504 arrangements have the green and the pink card next to each other. How many do not? 504-28= 476
 
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ok i thought we also had to find out the solutions of inhow many p was there
 
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P(Judy wins with the 2nd choice of cards)= P(0) + P(>=4)

P(0)= 5/25 [you can use a two way table to find out]
and P(>=4) = 10/25 [use your prob. distribution table from part (ii) ]
hence, 5/25*10/25=2/25

(v) i think a relevant alternative ans could be 2/5^n...(though this does not match the mark scheme, somehow).
ok...this i get it P(>=4) but why do we have to include P(0)?
 
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ok...this i get it P(>=4) but why do we have to include P(0)?
the ques. says that " If the score is 0 they play again"...and then in part (iv) they ask you to find the probabilty that Judy wins with the second choice of cards...hence you include the prob. of zero , times the prob. that judy wins..
hope tht helps
 
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the ques. says that " If the score is 0 they play again"...and then in part (iv) they ask you to find the probabilty that Judy wins with the second choice of cards...hence you include the prob. of zero , times the prob. that judy wins..
hope tht helps
ohh!! now i get it....i didn't read the question carefully...
thank u so much :)
 
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Can someone please explain to me how we find out if two variables are independent of each other?
 
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this one i can help u with 4b and 7ii but the other two m not sure
4b) 1 2 3 4 5 6
here u want the number of ways 4 and 5 are not next to each other, so u will have to find number of ways they are next to each other then subtract it from the number of ways the six numbers can be arranged in a line
so first number of ways all numbers can be arranged = 6! =720
now to find number of ways they are next to each other, u can take 4 and 5 as one unit..i mean
1 2 3 4 5 6 u can take it as one so that now u have 5 numbers to arrange that will be 5! and since 4 and 5 are different u can arrange them in 2 ways
so ur answer will be 6!-(5!*2) = 480
so prob =480/720 = 2/3
hope i was clear :)
 
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this one i can help u with 4b and 7ii but the other two m not sure
4b) 1 2 3 4 5 6
here u want the number of ways 4 and 5 are not next to each other, so u will have to find number of ways they are next to each other then subtract it from the number of ways the six numbers can be arranged in a line
so first number of ways all numbers can be arranged = 6! =720
now to find number of ways they are next to each other, u can take 4 and 5 as one unit..i mean
1 2 3 4 5 6 u can take it as one so that now u have 5 numbers to arrange that will be 5! and since 4 and 5 are different u can arrange them in 2 ways
so ur answer will be 6!-(5!*2) = 480
so prob =480/720 = 2/3
hope i was clear :)
thankyou :)
 
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