There are three identical desks, each with two drawers. In one desk, both drawers contain gold coins. In another desk, both drawers contain silver coins. In the third desk, one drawer holds a silver coin and the other drawer holds a gold coin. You choose a desk at random, and open one of the drawers at random. If you see a gold coin, what is the probability that the other drawer in that desk also contains a gold coin?

P ( B / A ) = 1/3

Step-by-step explanation:

Solution:-

You select a drawer and select a coin from that drawer.

Let A denote the event that the coin that you select is silver.

Let B denote the event that the other coin in the drawer is gold.

You are equally likely to pick any coin, so:

                 P ( A ) = 3 silver coins / 6 coins in total = 0.5

There are 6 different ways you could have selected a coin. Only one of them results in you selecting a silver coin AND the other coin in the drawer being gold, so:

                 P ( A & B ) = 1 desired / 6 total coins = 1/6

By the definition of conditional probability:

                P ( B / A ) = P ( A & B ) / P ( A )

                                 = 1/6 / 0.5

                                 = 1 /3

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)