Question #326183

Suppose that an examination consists of six true and false questions, and assume that


a student has no knowledge of the subject matter. The probability that the student will


guess the correct answer to the first question is 30%. Likewise, the probability of


guessing each of the remaining questions correctly is also 30%.


a. What is the probability of getting more than three correct answers?


b. What is the probability of getting at least two correct answers?


c. What is the probability of getting at most three correct answers?


d. What is the probability of getting less than five correct answers

1
Expert's answer
2022-04-14T08:54:07-0400

Xnumberofcorrectanswers,XBin(6,0.3)a:P(X>3)=i=46P(X=i)=i=46C6i0.3i(10.3)6i==150.340.72+60.350.71+0.36=0.07047b:P(X2)=1P(X<2)=1P(X=0)P(X=1)==10.7660.30.75=0.579825c:P(X3)=1P(X>3)=10.07047=0.92953d:P(X<5)=1P(X5)=1P(X=5)P(X=6)=160.350.710.36=0.989065X-number\,\,of\,\,correct\,\,answers, X\sim Bin\left( 6,0.3 \right) \\a:\\P\left( X>3 \right) =\sum_{i=4}^6{P\left( X=i \right)}=\sum_{i=4}^6{C_{6}^{i}\cdot 0.3^i\cdot \left( 1-0.3 \right) ^{6-i}}=\\=15\cdot 0.3^4\cdot 0.7^2+6\cdot 0.3^5\cdot 0.7^1+0.3^6=0.07047\\b:\\P\left( X\geqslant 2 \right) =1-P\left( X<2 \right) =1-P\left( X=0 \right) -P\left( X=1 \right) =\\=1-0.7^6-6\cdot 0.3\cdot 0.7^5=0.579825\\c:\\P\left( X\leqslant 3 \right) =1-P\left( X>3 \right) =1-0.07047=0.92953\\d:\\P\left( X<5 \right) =1-P\left( X\geqslant 5 \right) =1-P\left( X=5 \right) -P\left( X=6 \right) =1-6\cdot 0.3^5\cdot 0.7^1-0.3^6=0.989065


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