Q::Two 16 seater mini buses.A and B, are awaiting at a station. A and B have already had 11 and 12 seats occupied respectively. Each passenger selects A with probability 4/7 and selects B with probability 3/7.Find the probability that A is filled up first.
Hint:
The probability distribution requires 5 more passengers for A while there are 3, 2, 1 or 0 more passengers in B. The orders of boarding A and B are not important except the last one, who must board bus A. Therefore (n-1)Cr is used.
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(presentation for answer (e.g. "Let ... be ...") is omitted)
Bus A: 16-11=5
bus B: 16-12=4
PE=P(5A)+P(4A1B+last one is A)+P(4A2B+1A)+P(4A2B+1A)+P(4A3B+1A)
PE=(4/7)^5+5C4*(4/7)^5*(3/7)+6C4*(4/7)^5*(3/7)^3+7C4*(4/7)^5*(3/7)^3
PE=0.5272