skamal7 wrote:
WHy in both winning combination we are calculating for GG only once .
May be on first die 5 and second die one or on first die one and second die 5...These can combinations can also occur na? ..WHy we are not considering this scenario?
Hi skamal7,
Consider the following example that will explain better than any theoretical information.
You say that G,G should be counted twice, so the possible combinations are:
G,G=1/9
G,G=1/9
B,G=2/9
G,B=2/9
B,B=4/9
[ also if your method is correct B,B should be counted twice =4/9 ]
don't you see anything odd? The sum of the probability of each case is greater than 1! \(\frac{1+1+2+2+4}{9}=\frac{10}{9}\)
[ if you count B,B twice it becomes \(\frac{14}{9}\) ]
Why does this happen?Let's look at the theory now
The formula to solve this problem is \((nCk)p^k*q^{(n-k)}\) where p=1/3 and q=2/3 and N are the dies and K are the good outcomes:
Case two good \((2C2)(\frac{1}{3})^2(\frac{2}{3})^0=\frac{1}{9}\)
Case one good one bad \((2C1)(\frac{1}{3})^1(\frac{2}{3})^1=\frac{4}{9}\)
Case two bad \((2C0)(\frac{1}{3})^0(\frac{2}{3})^2=\frac{4}{9}\)
Tot sum = \(\frac{1+4+4}{9}=1\)
Hope it's clear now, let me know