No, using BODMAS you do any multiplication first unless there are Brackets or Indices (Order), so their answer is correct.
To get zero you need to do something like:
(777+1)*2*3*4*5*6*7*8*9*0
or
(777+1*2*3*4*5*6*7*8*9)*0
HTH,
Andrew
PS still not done the rope in the lake problem - busy day at work!!!
deleted user 10.02.2005 04:45
I don't understand the Question no. 3 :S Can anybody tell me what I should do to get the right answer
What's the answer of question 6???
Question:
A deck of cards contains 16 black cards and 4 red cards.
What is the probability of getting all 4 red cards if you randomly choose 4 cards from the deck?
1 in ?
I thought it was 100 / ( (4/20) * (3/19) * (2/18) * (1/17) ) and that's 1 in 484500... But that's not good.
3) What is the probability of getting "heads" on every throw when flipping a coin seven times?
1 in ?
The probability of a head on the toss of a coin is 1/2.
If we throw it a second time the results are independent of each other so the probability is 1/2 again.
But, the probability of two heads in a row is 1/2 * 1/2 = 1/4 (another way to look at this is that you can get HH, HT, TH, TT and each is equally likely so that there is a 1 in 4 chance of two heads in a row)
Each time the coin is tossed, the number of possible outcomes doubles (ie, 3 tosses, 8 outcomes:- HHH, HHT, HTH, THH, HTT, TTH, THT, TTT), and there is always only 1 way you can get only heads.
So in 7 tosses, you need to work out how many possible outcomes there are if you keep doubling.
HTH,
Andrew
the answers to ANY probability questions are not 1 in ????
they are just the "in" element - ie - if your answer is 1 in 25
then the answer to enter is 25