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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?

Respuesta :

MrRoyal

Answer:

[tex]P(R_2Y_2C_2B_2) = 0.03845[/tex]

Step-by-step explanation:

Given

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

[tex]Turns = 8[/tex]

Required

[tex]P(R_2Y_2C_2B_2)[/tex]

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

[tex]P(R) = P(Y) = P(C) = P(B) = \frac{1}{4}[/tex]

From [tex]P(R_2Y_2C_2B_2)[/tex], we have:

[tex]R = Y = C = B = 2[/tex]

[tex]Total = 8[/tex]

So, the total arrangement of the 8 balls is:

[tex]Arrangement = \frac{8!}{2!2!2!2!}[/tex]

[tex]Arrangement = \frac{40320}{2*2*2*2}[/tex]

[tex]Arrangement = \frac{40320}{16}[/tex]

[tex]Arrangement = 2520[/tex]

The individual probability of each ball, when put together is

[tex]Probability = P(R)^2 * P(Y)^2 * P(C)^2 * P(B)^2[/tex]

[tex]Probability = (1/4)^2 *(1/4)^2 *(1/4)^2 *(1/4)^2[/tex]

[tex]Probability = (1/16) *(1/16) *(1/16) *(1/16)[/tex]

[tex]Probability = \frac{1}{65536}[/tex]

Lastly:

[tex]P(R_2Y_2C_2B_2) = Arrangement * Probability[/tex]

[tex]P(R_2Y_2C_2B_2) = 2520 * \frac{1}{65536}[/tex]

[tex]P(R_2Y_2C_2B_2) = \frac{2520 }{65536}[/tex]

[tex]P(R_2Y_2C_2B_2) = 0.03845[/tex]