Suppose you are rolling a die for 4 times.
Let x be the random variable finding how many times you will get a ‘3’.
a. List all possible values for x.
b. Find the probability distribution of x.
a. Possible values for X = 0,1,2,3,4
b. The sample space of equally likely outcomes is
"P(3) = \\frac{34}{36} \\\\\n\np=0.055 \\\\\n\nq = 0.945 \\\\\n\nn = 4 \\\\\n\nP(X=0) = C^4_0 (0.055)^0(0.945)^{4-0} \\\\\n\n= \\frac{4!}{0!(4-0)!} \\times 1 \\times 0.7974 \\\\\n\n= 0.7974 \\\\\n\nP(X=1) = C^4_1 (0.055)^1(0.945)^{4-1} \\\\\n\n= \\frac{4!}{1!(4-1)!} \\times 0.055 \\times 0.8439 \\\\\n\n= 0.1856 \\\\\n\nP(X=2) = C^4_2 (0.055)^2(0.945)^{4-2} \\\\\n\n= \\frac{4!}{2!(4-2)!} \\times 0.003025 \\times 0.8930 \\\\\n\n= 0.0162 \\\\\n\nP(X=3) = C^4_3 (0.055)^3(0.945)^{4-3} \\\\\n\n= \\frac{4!}{3!(4-3)!} \\times 0.00016638 \\times 0.945 \\\\\n\n= 0.0006289 \\\\\n\nP(X=4) = C^4_4 (0.055)^4(0.945)^{4-4} \\\\\n= \\frac{4!}{4!(4-4)!} \\times 9.15 \\times 10^{-6} \\times 1 \\\\\n= 9.15 \\times 10^{-6}"
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