To answer this question, we must first note that the probability of x successes in n trials is:
P = nCx*[tex] p^{x} [/tex]*[tex] q^{n-x} [/tex] where p and q are the
probabilities of success and failure respectively.
In this case:
p = 33% = .33
q = .67
n = 10
x = 6
Calculating [tex] p^{x} [/tex]*[tex] q^{n-x} [/tex]:
[tex] p^{x} [/tex]*[tex] q^{n-x} [/tex]=0.00026
Now the number of combinations of x in n trials is:
nCx = [tex] \frac{n!}{x!(n-x)!} [/tex]
Using the calculator to calculate for nCx from x= 1 to 6:
10C1 = 1
10C2 = 45
10C3 = 120
10C4 = 210
10C5 = 252
10C6 = 210
total nCx = 838
Substituting known values to the P equation:
P(6 or below are physics majors) = 838 *
0.00026
P(6 or below are physics majors) = 0.218
