# Test with 4 true/false questions

A student is writing a test with 4 true/false questions. The student wants to know the theoretical probability of passing the test by guessing. This means that 2 or more of the 4 questions are answered correctly by guessing. Determine this probability and show all your work.

I don't know what to do next and how to approach it.

SO since there is 2 options and 4 questions it would be 0.5^2 But It says 2 or more, which is why I am confused.

• Is it exactly two, or atleast two hearts? – Tavish May 16 '20 at 20:03
• exactly two hearts – Salam May 16 '20 at 20:03
• You just changed the question? Why? – Tavish May 16 '20 at 20:04
• I noticed I was asking the wrong one. – Salam May 16 '20 at 20:06

The student can get $$2$$ questions right and $$2$$ wrong, or $$3$$ right and $$1$$ wrong, or all $$4$$ right. For him to get all $$4$$ right, the probability is $$0.5^4$$.

For him to get $$3$$ right and $$1$$ wrong, there are $$4$$ ways to pick which question he gets wrong, then $$0.5$$ probability that he gets it wrong, then $$0.5^3$$ probability he gets the other $$3$$ questions right.

Finally, for him to get $$2$$ right and $$2$$ wrong, there are $$\binom{4}{2} = 6$$ ways to pick which $$2$$ questions he gets right, $$0.5^2$$ probability he gets them right, and then $$0.5^2$$ probability he gets the other $$2$$ questions wrong.

Therefore, the total probability is $$0.5^4 + 4 \cdot 0.5^3 \cdot 0.5 + 6 \cdot 0.5^2 \cdot 0.5^2 = 0.6875$$.

Note: Considering this sort of problem in the general case, where $$4$$ is replaced by $$n$$, $$2$$ is replaced by $$k$$, and $$0.5$$ is replaced by $$p$$, leads to the idea of a binomial random variable (or a random variable with the binomial distribution). You can look at http://onlinestatbook.com/2/probability/binomial.html.

This can be done through the use of Binomial Distribution. We want either $$2$$ successes, or $$3$$, or $$4$$:

$${4\choose 2} \cdot (0.5)^4 + {4\choose 3}\cdot (0.5)^4 + {4\choose 4} \cdot (0.5)^4 = \frac{11}{16}$$

Edit:

An easier way is to do the following. First, calculate $$P(0 \ \text{right} ) = (0.5)^4$$. Then, calculate $$P(1 \ \text{right}) = 4\times (0.5)^4$$. Here, the factor of $$4$$ comes from the fact that the student could either get the first question right, or the second etc. and we need to make sure that the other three are answered wrongly, which is the reason for multiplying $$0.5$$ four times. Now, $$P(\ge 2) = 1-P(\lt 2) \\ =1-(P(0) + P(1)) \\ = 1-( (0.5)^4 + 4\times (0.5)^4) \\=\frac{11}{16}$$

• I have not taken binomial distribution and so I would have to solve it differently – Salam May 16 '20 at 20:13
• @Salam I added an alternate solution. – Tavish May 16 '20 at 20:22