At a round table, n persons are seated on n chairs. The probability that two friends from same college are sitting next to each other, is:
Step 1 — Total circular arrangements: Seating \(n\) distinct persons around a round table gives \((n-1)!\) arrangements.
Step 2 — Favourable arrangements: Treat the two friends as one block. The block plus the other \((n-2)\) persons gives \((n-2)!\) circular arrangements, and the two friends can be arranged within the block in \(2!\) ways. Favourable count \(=2\,(n-2)!\).
Step 3 — Probability:
\[P=\frac{2\,(n-2)!}{(n-1)!}=\frac{2}{n-1}\]
Hence the required probability is \(\dfrac{2}{n-1}\).
The chance that a software engineer will debug an error X correctly is 80%. The chance that software with the error X will crash after the correct debugging is 30%. The chance of crash of the software by wrong debugging is 70%. A software with the error X crashed. The probability that its error was debugged correctly is:
A six faced die is a biased one. It is thrice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even is
A coin is biased so that a head is twice as likely to occur as a tail, if the coin is tossed three times, what is the probability of getting exactly two tails?
From two well shuffled pack of cards what is the probability of getting one Jack from the first one and a King from the second?
If A is an event of getting 13 by throwing two unbiased six-faced dice, then A is called
One summer, Peter visits 4 villages (A, B, C and D) in a random order. Find the probability that he visits (i) A before B (ii) A before B and B before C respectively?