A car is parked by a driver amongst 25 cars in a row, not at either end. When he returns he finds that 10 places are empty. The probability that both the neighbouring places of drivers car are vacant, is???Pls explain how???

Question

jitender lakhanpal · Answer

Dear,

answer is C(23,8) / C (25,10)

for solution

please find the attachment

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Kunal Upadhyay · Answer

Since the owner finds on return that exactly 10 of the 25 Places(inckuding owner’s car) are occupied, the exhaustive no. of cases for such an arrangement is (25-1) C( 10-1) = 24 C 9 [ since the remaining 9 cars to be parked in remaining 24 places] Let E denote the event that both neighbouring places to owner’s car are empty. This require thr remaning 9 cars to be parked in remainig (25-3)=22 Places and hence the no of cases favourabke to E is 22 C 9 . Hence, P(E) = 22 C 9 / 24 C 9 = 35/92 = 0.38043 Ans.

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A car is parked by a driver amongst 25 cars in a row, not at either end. When he returns he finds that 10 places are empty. The probability that both the neighbouring places of drivers car are vacant, is???Pls explain how???

A car is parked by a driver amongst 25 cars in a row, not at either end. When he returns he finds that 10 places are empty. The probability that both the neighbouring places of drivers car are vacant, is???Pls explain how???

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A car is parked by a driver amongst 25 cars in a row, not at either end. When he returns he finds that 10 places are empty. The probability that both the neighbouring places of drivers car are vacant, is???Pls explain how???

A car is parked by a driver amongst 25 cars in a row, not at either end. When he returns he finds that 10 places are empty. The probability that both the neighbouring places of drivers car are vacant, is???Pls explain how???

2 Answers

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