There is a 2 of 6 percent chance that one them might land on three if you add one extra dice it would be 3 of 6 percent chance and so on.
Hope this Helped!
;D
Brainliest??
Here we might have to find p(v intersection w) and for that we use the following formula
p(v U w) = p(v)+p(w)-p(v intersection w)
And it is given that p(v) =01.3 , p(w) = 0.04 and p(v U w ) = 0.14 .
Substituting these values in the formula, we will get
0.14 = 0.13 +0.04 -p(v intersection w)
p(v intersection w) =0.13 +0.04 -0.14 = 0.03
So the required answer of the given question is 0.03 .
15p^2q^2 because you multiply the coefficients and the vaiables. 3x5 is 15, pxp is p^2, and qxq is q^2
The set of fractions between 1 and 2 are infinite.
The answer would be 32/5 or 6 2/5 or 6.4
Hope this helps
Have a great day/night
