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A PC has two color options, white and gray. A customer demands the white PC with a probability of 0.3. A seller of these PCs has three of each color in stock, although this is not known to the customers. Customers arrive and independently order these PCs.

Find the probability that all of whites are ordered before all of the grays.

My attempt

If all whites are ordered before grays it means that the 6th order will always be gray Therefore I found the probability that the 6th demand would be gray

X=number of trials

R=number of R-th success

Formula Combination of $$\binom{X-1}{R-1} \cdot p^R \cdot (1-p)^{X-R}$$

Applying formula: $$\binom{5}{2} \cdot 0.7^3\cdot 0.3^3=0.0926.$$

However the answer in the answer key is $0.16308.$

Can someone tell me where I went wrong, Thanks!

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Let's distinguish some cases.

  • First case: The white PC's are bought in the first 3 sales. The probability for this to happen is: $$p_1 = 0.3^3 = 0.027.$$

  • Second case: The white PC's are bought in the first 4 sales. Notice that the last bought PC is white (otherwise, the "experiment" stops at the first 3 sales). The probability for this to happen is: $$ p_2 = \binom{3}{2} \cdot 0.3^3 \cdot 0.7 = 0.0567.$$

  • Third case: The white PC's are bought in the first 5 sales. Notice again that the last bought PC must be white. Thus: $$ p_3 = \binom{4}{2} \cdot 0.3^3 \cdot 0.7^2 = 0.07938.$$

There is no other case. If we add up all the probabilities (because the events are mutually exclusive) we have that: $$p_1 + p_2 + p_3 = 0.16308.$$

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