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Imagine that you do not understand anything in this class. You would have to randomly guess the answer for the above 5 problems. What is the probability that you can get at least two out of five right?
  1. $\sum_{k=2}^{5}\left(\frac{1}{4}\right)^{k}$
  2. $\sum_{k=2}^{5}\binom{5}{k}\left(\frac{1}{4}\right)^{k}\left(\frac{3}{4}\right)^{5-k}$
  3. $\left(\frac{1}{4}\right)^{2}\left(\frac{3}{4}\right)^{3}$
  4. $\sum_{k=2}^{5} \frac{5!}{k!}\left(\frac{1}{4}\right)^{k}\left(\frac{3}{4}\right)^{5-k}$
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Every question have 4 options and any one of the option is correct.

P(guessed option is correct) = 1 / 4

P(guessed option is not correct) = 3 / 4

Probability(Atleast two out of 5 are correct) = P(2 are correct) + P(3 are correct) + P(4 are correct)

+ P(5 are correct)

 


P(2 are correct) = $_{2}^{5}\textrm{C} \left ( 1/4 \right )^{2} \left ( 3/4 \right )^{3}$

P(3 are correct) = $_{3}^{5}\textrm{C} \left ( 1/4 \right )^{3} \left ( 3/4 \right )^{2}$

P(4 are correct) = $_{4}^{5}\textrm{C} \left ( 1/4 \right )^{4} \left ( 3/4 \right )^{1}$

P(5 are correct) = $_{5}^{5}\textrm{C} \left ( 1/4 \right )^{5} \left ( 3/4 \right )^{0}$


 

Probability(Atleast two out of 5 are correct) =  

$_{2}^{5}\textrm{C} \left ( 1/4 \right )^{2} \left ( 3/4 \right )^{3}$ + $_{3}^{5}\textrm{C} \left ( 1/4 \right )^{3} \left ( 3/4 \right )^{2}$ + $_{4}^{5}\textrm{C} \left ( 1/4 \right )^{4} \left ( 3/4 \right )^{1}$ + $_{5}^{5}\textrm{C} \left ( 1/4 \right )^{5} \left ( 3/4 \right )^{0}$

=  $\sum_{k=2}^{5} \binom{5}{k} \left ( 1/4 \right )^{k} \left ( 3/4 \right )^{5-k}$


 

Option B, is Correct Answer.

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