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  • The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is A.7/64

The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is
A.\frac{7}{64}

B.\frac{7}{128}

C.\frac{45}{1024}

D.\frac{7}{41}

Answers (1)

We know that in the examination, we have only two option True and False.

Therefore, the probability of getting True is, p= 1/2

And, probability of getting False is, q=1/2

The total number of Answer in examination, n=10
The probability of guessing correctly at least 8 it means r=8,9,10

As we know, the probability distribution \mathrm{P}(\mathrm{X}=\mathrm{r})={ }^{n} \mathrm{C}_{\mathrm{C}}(\mathrm{p})^{r} \mathrm{q}^{\mathrm{n}-\mathrm{r}}
\mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{P}(\mathrm{r}=8)+\mathrm{P}(\mathrm{r}=9)+\mathrm{P}(\mathrm{r}=10)
\\ =^{10} \mathrm{C}_{8}\left(\frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)^{10-8}+^{10} \mathrm{C}_{9}\left(\frac{1}{2}\right)^{9}\left(\frac{1}{2}\right)^{10-9}+^{10} \mathrm{C}_{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^{10-10}$ \\$=\left(\frac{10 !}{8 ! 2 !}+\frac{10 !}{9 !}+1\right)\left(\frac{1}{2}\right)^{10}$ \\$=[45+10+1]^{\left(\frac{1}{2}\right)^{10}}$ \\$=\frac{56}{1025}=\frac{7}{128}
Hence, Probability is  \frac{7}{128}

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