benjaminjacob98 wrote:
why doesn't it work when we calculate (1-probability of both balls being red) 1-6/15*5/14
because when you mltiply by 5/14, this implies that your second pick is red given the first one has been red
we cannot select red at all
it's (1-6/15)*(1-6/14)=9/15*8/14=24/70 simplied to 12/35
Also, we can select two slots: ____ ____ ----> 9 (not red for the first slot)*8 (not red for the second slot). Since this is not permutation, we must divide by a factor of 2 or the number of slots. Hence 9*8/2=36 is the number of favorable outcomes.
In total there are 15C2 outcomes or 15!/(2!*13!)=15*14/2=105
The answer must be 36/105 and this is the same as 12*3/35*3 reduced to 12/35