The probability of a student speaking spanish is $30\%$. If we select $3$, what are the chances of at least one of them speaking Spanish?

Question

In a school the probability of a student speaking Spanish is $30\%$. If we select $3$ random students, what are the chances of at least one of them speaking Spanish?

So, I saw this question and tried to solve it, seemed like an easy question but I was wrong and still not sure why.

Apparently the correct answer is $65.7\%$ ($100\% - 70\% \cdot 70\% \cdot 70\%$)

My first attempt at solving I figured $90\% = 3 \cdot 30\%$ (I just need $1$ student to know spanish, and I have $3$ "chances").

What is the corrent approach for this?

I created this question but for some reason it created in a non-registered account, sorry for the duplicate.

Also, I used the word odds, but I meant probability, english is not my first language and the question was written in portuguese. Sorry for confusion

Answer 1

Calculate $1$ minus the probability of the complementary event:

• The probability that a student speaks Spanish is $0.3$
• The probability that a student does not speak Spanish is $1-0.3$
• The probability that all $3$ students do not speak Spanish is $(1-0.3)^3$
• The probability that not all $3$ students do not speak Spanish is $1-(1-0.3)^3$

Answer 2

The complementary event of $A = $"at least one student speaks Spanish" is that $\bar A =$None (of the three) speaks Spanish. To calculate the chance that no student speaks Spanish, it has to be the case that the three students do not speak Spanish which is $(1-.3)\times (1-.3)\times (1-.3)$ . Hence, $$P(A) = 1-P(\bar A) = 1-(1-.3)^3 = 0.657$$

The reason your answer is wrong is because you (almost) calculated the chance of the event "exactly one student speaks Spanish" not "at least on speaks Spanish". To calculate the chance that exactly one student speaks Spanish, first you need to count the ways there are that one of the three speaks it and there are $\binom{3}{1} = 3$ ways. The other two do not speak it. Hence, there is a $$\binom{3}{1}(.30)^1(1-.3)^2 = 0.63$$ chance that exactly one speaks it. Similarly, you can calculate the chance of getting 2 or 3.

The event "at least one" is the same as 1 or 2 or 3 speaks Spanish. Since the events are disjoint, you can add up the probabilities and so , $$P(A) = \sum_{k = 1}^3\binom{3}{k}(.3)^k(1-.3)^{3-k} = 0.657.$$