Answer to Question #147155 in Statistics and Probability for Vladimyr Lubin

Question #147155
(5.2.17) Assume that random guesses are made for seven multiple choice questions on an SAT​ test, so that there are n=7 ​trials, each with probability of success​ (correct) given by p=0.35. Find the indicated probability for the number of correct answers.
Find the probability that the number x of correct answers is fewer than 4.
P(X<4)=? (Round to four decimal)
1
Expert's answer
2020-11-30T12:16:46-0500

n = 7

P(success) = 0.35

The number of successes among a fixed number of independent trials follows binomial distribution. Lets evaluate the definition of binomial probability at m = 0, 1, 2, 3:


P(X=m)=C(n,m)pm(1p)nmP(X=m)=C(n,m)\cdot p^m\cdot(1-p)^{n-m}

P(X=0)=C(7,0)0.350(10.35)70=7!0!(70)!0.3500.657=0.049P(X=0)=C(7,0)\cdot 0.35^0\cdot(1-0.35)^{7-0}=\frac{7!}{0!(7-0)!}\cdot0.35^0\cdot0.65^7=0.049

P(X=1)=C(7,1)0.351(10.35)71=7!1!(71)!0.3510.656=0.1848P(X=1)=C(7,1)\cdot 0.35^1\cdot(1-0.35)^{7-1}=\frac{7!}{1!(7-1)!}\cdot0.35^1\cdot0.65^6=0.1848

P(X=2)=C(7,2)0.352(10.35)72=7!2!(72)!0.3520.655=0.2985P(X=2)=C(7,2)\cdot 0.35^2\cdot(1-0.35)^{7-2}=\frac{7!}{2!(7-2)!}\cdot0.35^2\cdot0.65^5=0.2985

P(X=3)=C(7,3)0.353(10.35)73=7!3!(73)!0.3530.654=0.2679P(X=3)=C(7,3)\cdot 0.35^3\cdot(1-0.35)^{7-3}=\frac{7!}{3!(7-3)!}\cdot0.35^3\cdot0.65^4=0.2679


P(X<4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)P(X<4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)

P(X<4)=0.049+0.1848+0.2985+0.2679=0.8002P(X<4)=0.049+0.1848+0.2985+0.2679=0.8002


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