A learning experiment requires a rat to run a maze (a network of pathways) until it locates one of three possible exits. Exit 1 presents a reward of food, but exits 2 and 3 do not. (If the rat eventually selects exit 1 almost every time, learning may have taken place.) Let Yi denote the number of times exit i is chosen in successive runnings. For the following, assume that the rat chooses an exit at random on each run. Required:
Find the probability that n = 6 runs result in Y1=3,Y2=1, and Y3=2

Find the probability that n = 6 runs result in Y1=3,Y2=1, and Y3=2

=0.823

Step-by-step explanation:

Multinomial distribution

p(y1;y2,y3…..yk) = (n!/y1!,y2!,y3!…y4!) x p1^y1 x p2^y2 x p3^y3 x pk^yk)  

The rat chooses of the three exists at random:

p = 1/3

p(3,2,1) = 6! / 3! 1!  2! (1/3)^3 (1/3)^1 (1/3)^2

              =20/243

              =0.823

unpaid balance is : $149.99-$50 = $99.99 new balance = 149.99+2.62-50 = $102.61 finance charge = 99.99- 0.21/12= $1.75 new balance = 99.99+1.75 = $101.74 average daily balance method average daily balance= 3399.70/30 = $113.32 finance charge = $ 0.21/12ă—113.32 = $1.98 new balance = $149.99-50+1.98 = $101.97

560

step-by-step explanation:

giggidy