Answer:
When a card is chosen at random with replacement five times, X is the number of times a prime number is chosen. Here the card is chosen with replacement. This implies probability for choosing a prime number remains the same as the previously drawn card is replaced.
The sample space= {1,2,3,4,5,6,7,8,9,10}
Prime numbers = {2,3,5,7}
Prob for drawing prime number = 4/10 = 0.4
is the same when replacement is done.
Also there are two outcomes either prime or non prime. Hence in this case, X the no of times a prime number is chosen, is binomial with p =0.4 and q = 0.6 and n=5
When a card is chosen at random without replacement three times, X is the number of times an even number is chosen.
Prob for an even number = 0.5
But after one card drawn say odd number next card has prob for even number as 5/9 hence each draw is not independent of the other. Hence not binomial.
When a card is chosen at random with replacement six times, X is the number of times a 3 is chosen.
Here since every time replacement is done, probability of drawing a 3 remains constant = 1/10 = 0.3
i.e. each draw is independent of the other and there are only two outcomes , 3 or non 3. Hence here X is binomial.
When a card is chosen at random with replacement multiple times, X is the number of times a card is chosen until a 5 is chosen
Here X is the number of times a card is chosen with replacement till 5 is chosen. This is not binomial. Here probability for drawing nth time correct 5 is P(non 5 in the first n-1 draws)*P(5 in nth draw) = 0.1^(n-1) (0.9)
Because nCr is not appearing i.e. 5 cannot appear in any order but only in the last draw, this is not binomial.
Step-by-step explanation:
. Simplifying a bit gives us 
and 
. Raising that number inside the parenthesis to the 15th power gives us 
. Multiplying to finish means that A(t) = $317.22