need help. show the steps

Question

A psychic network recieved 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 psychics agreed to take part in the following expierment. 5 different cards are shuffled and one is chosen at random. the psychic will then try top identify which card was drawn without seeing it. Assume that the expierment was repeated 45 times and the results of any two expeirments are independent of one another. If we assume that the psychic is fake. A) How many of the 45 cards do we expect the psychic to guess correctly. B)FInd the probability that they will guess at least three correctly

Answer 1

If the psychic is fake, then they can only identify the card at random; since there are 5 possible cards, of which only 1 is right, this means that their chance of guessing correctly is 1/5, or .2.  Since there are two possible outcomes with fixed probabilities per trial, this is a binomial distribution problem.

 

A)  Given n trials (here 45) and a probability p of a 'successful' outcome on any one trial (here .2), the expected number of successful outcomes is simply np.  Here, that's 45*.2, or 9.  This should make some degree of sense - if they get the right card by chance a fifth of the time, then you should expect them to guess a fifth of those 45 cards correctly, and a fifth of 45 is 9.

 

B)  To guess "at least three" correctly implies the probability of them guessing exactly 3 correctly, plus the probability of them guessing exactly four correctly, plus that of exactly five, all the way up to exactly 45 correct.  You can calculate each of these with the binomial distribution, but it's much simpler to go the other way.  The probability of every outcome must add up to 1, so we can save ourselves some time by calculating the probability that they will NOT guess at least three correctly (i.e., that they guess exactly 0, 1, or 2 correctly), and just subtract that from 1.

 

The probability of any exact number of successful outcomes k with a binomial distribution can be calculated as:

 

n!/(k!(n-k)!) * p^k(1-p)^n-k

 

With n and p again being the total number of trials (45) and the probability of a successful trial (.2) respectively.

 

So, the probability of the psychic guessing exactly 0 cards successfully is:

 

45!/45!0! * .2⁰*.8⁴⁵ = .8⁴⁵ = .000043

 

Of exactly 1 card:

 

45!/44!1! * .2¹*.8⁴⁴ = 45*.2*.8⁴⁴ = .00049

 

Of exactly 2 cards:

 

45!/43!2! * .2²*.8⁴³ = (45*44)/2*.2²*.8⁴³ = .002695

 

Which, when added up, give .003228.  As stated before, this is the probability that they will NOT guess at least 3 correctly.  To get the probability that they will, we just subtract it from 1:

 

1 - .003228 = .99677.

 

Well over a 99% chance.

Answer 2

This is essentially a binomial distribution with n=45 and p=.2

 

The expected number number of correct guesses is the mean of the distribution = np =9.

 

Again since it is binomial can you figure out how to get P(at least 3 right)?

 

 

Any questions comment back.