ok, i am back. again a PS. Time yourself. the fastest and correct answer wins. Imagine a parking lot with 500 cars with license plates numbered from 001 to 500 and no two cars having the same license plate number. At 6 pm, they all leave…
ok, i am back. again a PS.
Time yourself. the fastest and correct answer wins.
Imagine a parking lot with 500 cars with license plates numbered from 001 to 500 and no two cars having the same license plate number. At 6 pm, they all leave the lot one by one. What is the probability that the license plate numbers of the first four cars to leave are in increasing order of magintude?
1/8
1/16
1/24
1/64
1/128
1/64 ?
Can't exaplain.. I use sixth sense for prob 😞
The answer is 1/4! = 1/24.
The logic is that if you randomly select any 4 cars(numbers), the number of way
u can arrange these numbers in 4! ways and only 1 way out of them will
be the ascending one.
Nagging is right. the answer is 1/24
Sincerely
Aejaz
The question reduces down to the probability of 4 cars being chosen in ascending order. So uou have four cars. Only one arrangement is the ascending order. total arrangements possible are 4!=24. Therefore, the required probability is 1/4! that is 1/24.
Three numbers are chosen at random w/o replacement from (1,2,3...10). The probability that the minimum no is 3 or the maximum no is 7 is...
(please give the detailed solution)