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[sksuriza]

plz teach me how to solve q 5a2 of oct/nov 09

 

[ahmed t]

which variant??

 

[sksuriza]

61

 

[hallaballa]

i've posted it in maths 62

 

[sksuriza]

oct nov 10..61..q 6 part 4 n5????

 

[remz]

oct nov 10..61..q 6 part 4 n5????

I think part 4 would be solved liked this..

3 different color so we have to choose like this

2 R,B,G = 4!/2!
2 B,G,R = 4!/2!
2 G,B,R = 4!/2!
Add all of them which is equal to 36 and then multiply it by 6C3 ( Because you are choosing three different colors from six different colors) . Now why did i add 4!/2! three times? This is because there are three combinations of (2R,B,G). If you want , you could have simply done 4!/2! x 3 and then multiply it by 6C3. The answer is 720.

Trying to work on the last part in the meanwhile!

 

[sksuriza]

oct nov 10...63....plz explain ans of q 5 part 3

 

[Amna]

from the table, you can see that the total number of students is 194. out of these, (30 + 30)= 60 take between 7 to 10 minutes.
one student takes 7-10 minutes from the two students chosen at random. number of ways the one student can be chosen is 2C1
also note that this case is without replacement.
now P (student takes 7-10 minutes) = 60/194
and P (student takes less than 7 or greater than 10 minutes) = (194-60)/193 [193 because it's without replacement]
so we get P(one student takes 7-10 minutes) = 2C1 x 60/194 x 134/193 = 0.429