Question

On each turn of the knob, a gumball machine is equally likely to dispense a red, yellow, green or blue gumball, independent from turn to turn. After eight turns, what is the probability P[R2Y2C2B2] that you have received 2 red, 2 yellow, 2 green and 2 blue gumballs?

Answer 1

Answer:

$P(R_2Y_2C_2B_2) = 0.03845$

Step-by-step explanation:

Given

$Balls = 4$ --- [R, Y, C and B]

$Turns = 8$

Required

$P(R_2Y_2C_2B_2)$

Since each of the 4 balls are equally likely, their probability is:

$P(R) = P(Y) = P(C) = P(B) = \frac{1}{4}$

From $P(R_2Y_2C_2B_2)$, we have:

$R = Y = C = B = 2$

$Total = 8$

So, the total arrangement of the 8 balls is:

$Arrangement = \frac{8!}{2!2!2!2!}$

$Arrangement = \frac{40320}{2*2*2*2}$

$Arrangement = \frac{40320}{16}$

$Arrangement = 2520$

The individual probability of each ball, when put together is

$Probability = P(R)^2 * P(Y)^2 * P(C)^2 * P(B)^2$

$Probability = (1/4)^2 *(1/4)^2 *(1/4)^2 *(1/4)^2$

$Probability = (1/16) *(1/16) *(1/16) *(1/16)$

$Probability = \frac{1}{65536}$

Lastly:

$P(R_2Y_2C_2B_2) = Arrangement * Probability$

$P(R_2Y_2C_2B_2) = 2520 * \frac{1}{65536}$

$P(R_2Y_2C_2B_2) = \frac{2520 }{65536}$

$P(R_2Y_2C_2B_2) = 0.03845$