Anonymous

# probability question?

A psychic network received telephone calls last year from over 1.5 million people. A recent article attempts to shed some light onto the credibility of the psychic network. One of the psychic network's psychics agreed to take part in the following experiment. Five different cards are shuffled, and one is chosen at random. The psychic will then try to identify which card was drawn without seeing it. Assume that the experiment was repeated 25 times and that the results of any two experiments are independent of one another. If we assume that the psychic is a fake (i.e., they are merely guessing at the cards and have no psychic powers), find the probability that they guess at least three correctly.

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• 1 month ago

The only relevant numbers here are the number of cards to choose from, the number of trials, and the number of correct guesses.

They have a probability of 0.2 (1/5) of guessing the card correctly in any trial.  We would expect that they would guess the correct card 5 times in 25 trials, since 0.2 * 25 = 5 (with 3 guesses, 4 guesses, 6 guesses, 7 guesses being probable as well)

We just use a binomial distribution

P(n) = 25Cn * 0.2^n * (1 - 0.2)^n

P(0) = 25C0 * 0.2^0 * (1 - 0.2)^(25 - 0)

P(1) = 25C1 * 0.2^1 * (1 - 0.2)^(25 - 1)

P(2) = 25C2 * 0.2^2 * (1 - 0.2)^(25 - 2)

P(3) = 25C3 * 0.2^3 * (1 - 0.2)^(25 - 3)

....

P(24) = 25C24 * 0.2^24 * (1 - 0.2)^(25 - 24)

P(25) = 25C25 * 0.2^25 * (1 - 0.2)^(25 - 25)

If we sum it all up, P(0) + P(1) + ... + P(25), we'd get 1.  We want to sum up P(3) + P(4) + ... + P(25) instead.  Well, we could do all of that, or we could sum P(0) + P(1) + P(2) and subtract that from 1.  It's up to you.

25C0 * 0.2^0 * (1 - 0.2)^(25 - 0) =>

(25! / (0! * (25 - 0)!)) * 1 * 0.8^25 =>

1 * 1 * 0.8^25 =>

0.8^25

25C1 * 0.2^1 * (1 - 0.2)^(25 - 1) =>

(25! / (1! * (25 - 1)!)) * 0.2 * 0.8^24 =>

(25! / 24!) * 0.2 * 0.8^24 =>

25 * 0.2 * 0.8^24

25C2 * 0.2^2 * (1 - 0.2)^(25 - 2) =>

(25! / (2! * (25 - 2)!)) * 0.2^2 * 0.8^23 =>

(25 * 24 * 23! / (2 * 23!)) * 0.2^2 * 0.8^23 =>

25 * 12 * 0.2^2 * 0.8^23 =>

300 * 0.2^2 * 0.8^23

0.8^25 + 25 * 0.2 * 0.8^24 + 300 * 0.2^2 * 0.8^23 =>

(4/5)^25 + 25 * (1/5) * (4/5)^24 + 300 * (1/5)^2 * (4/5)^23 =>

(4/5)^25 + 25 * 4^(24) / 5^25 + 300 * 4^(23) / 5^25 =>

(4^25 + 25 * 4^24 + 300 * 4^23) / 5^25 =>

4^23 * (4^2 + 25 * 4 + 300) / 5^25 =>

(4/5)^(23) * (16 + 100 + 300) * (1/5)^2 =>

(4/5)^(23) * (416/25) =>

(4/5)^(23) * (1664/100) =>

16.64 * 0.8^23 =>

0.0982252228436886204448768

1 - 16.64 * 0.8^23 =>

0.9017747771563113795551232

There's a 90.18% chance of getting at least 3 correct.  Not exactly earth-shattering odds.

• 1 month ago

Psychics are often very religious and look upon their gift as a gift from god. They don't do any "now for my next trick" performances or show off in any way. They take their gift very seriously.