Suppose C1;C2;C3 are three didifferent biased coins, whose probability of heads equals 0.4, 0.5, and 0.2 respectively. Suppose coins are placed together in a box and you randomly picked a coin from the box. Flip the coin 10 times. Let A denote the event you randomly chose coin C1. Let B denote the event that you got exactly 4 heads out of the 10 coin flips. Compute the following probabilities:

P(A∩B)
P(B)
P(A|B)

Solution :

               = 0.0836

P(B)= P(coin chosen and 4 heads in 10 trails) + P(coin chosen and 4 heads in 10 trail) + P(coin chosen and 4 heads in 10 trail)

P(A|B) =

6

step-by-step explanation:

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube