+1124 Hi friends,
I have posted this a week ago, and got some responses. However, no-one has come back to me with regards to the problem I am having, so I am posting again in the hopes that someone will please explain this to me:
a Four digit code is requiered to open al ock. The code must be even numbered. May not contain 0 or 1. No repeats.
Calculate the probability that you will open the lock at first attempt if it is given that the code is greater than 5000 and the 3rd digit is 2.
They do give the answer, my problem is I do not understand the calculations, where do the numbers they use in the answer, come from?
(3*5*1*3)+(1*5*1*2)+(1*5*1*2)
=45+10+10
=65
(3*5*1*3)+(1*5*1*2)+(1*5*1*2)...this part...please explain this to me..?
+396 Hi Juriemagic,
I used a method like the one you are asking about, but my numbers were different.
I split the counting into four separate groups, first two digits odd - odd, odd - even, even - even or even - odd.
For odd - odd, there are three possibles for the first digit, 5, 7 or 9.
For the second digit there are again three possibles. There are four odd numbers 3, 5, 7, 9 but one of them has already been chosen, so three left to choose from.
The third digit is fixed and there are three possibles for the fourth digit.
That means that the total number of odd-odds is (3*3*1*3) = 27.
For the odd-even group it would be (3*3*1*2) = 18. The 2 at the end because one of the even numbers has been knocked out by the even second digit.
The other two, even-even and even-odd I'll leave you to work out, but the total of all four is 65.
As to the answer that you've posted, I think that the first digit in each bracket relates to the choices, odd, 6, 8, so 3, 1, 1.
If that's the case though, I don't understand where the 5 in the second position comes from.
+1124 Hi Tiggsy,
Finally somebody responds...thank you. I get what you say about how you calculate things...it really makes a lot of sense. Thank you so much. I will go ahead and try to do the last one....I will let you know..Thanks
+1124 Hi Tiggsy
okay, you gave for odds we have (3*3*1*3) =27..agrre
for odd/even, you calculated (3*3*1*2) = 18...agree
I have calculated evens....and got (3*2*1*1) = 6
and even/odd is (3*4*1*2) = 24
everything added together is 75?
+1124 Oh my goodness, yes off course, thanks a million...I also now finally understand this..
5324 , 5624 , 5724 , 5824 , 5924 , 6324 , 6524 , 6724 , 6824 , 6924 , 7324 , 7524 , 7624 , 7824 , 7924 , 8324 , 8524 , 8624 , 8724 , 8924 , 9324 , 9524 , 9624 , 9724 , 9824 , Total = 25
5326 , 5426 , 5726 , 5826 , 5926 , 7326 , 7426 , 7526 , 7826 , 7926 , 8326 , 8426 , 8526 , 8726 , 8926 , 9326 , 9426 , 9526 , 9726 , 9826 , Total = 20
5328 , 5428 , 5628 , 5728 , 5928 , 6328 , 6428 , 6528 , 6728 , 6928 , 7328 , 7428 , 7528 , 7628 , 7928 , 9328 , 9428 , 9528 , 9628 , 9728 , Total = 20
25 + 20 + 20 ==65 - This is the 2nd solution where they all end in 4, 6, 8. If you look at them carefully, they are very easy to understand. My problem with their solution is that I cannot get the denominator of 840!
+1124 Hi guest,
Thanks for your input, I'm looking at this now...no, the denominator is 840,Yep sorry, the 1st question (Which I did not post here), gave a different scenario, saying that a lock code must be even numbered, and may not contain digits 0 and 1. Digits may not be repeated. Calculate how many possible combinations are there to open the lock...This question here, is question 2, the second part...sorry about the confusion. So, we originally have 10 numbers. 0 an 1 may not be used, so we have 8 numbers, however the code must be even numbered, so we need to reserve one of the evens for the 4th spot, that leaves 7 numbers available...
First spot is one of 7 numbers, 2nd spot one of 6 numbers, 3rd spot one of 5 numbers and 4th spot, one of 4 numbers..
Thanks for your ijput, I'm looking at that now
