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I was thinking 1*1*1*2*2 = 4 out of 32, with LLLDD, LLLLL, LLLDL, LLLLD, with L as like and D as dislike. But if I can do LLLLD and LLLDL, why couldn't I do LDLLL or DLLLD? Any explanation would be appreciated.

EDIT: At least three (Sorry, forgot to mention)

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1  
Do you want the probability that he likes exactly three of the five, or at least three? –  Brian M. Scott Jan 29 '13 at 0:04
    
He sounds too picky, I doubt he will like any of them. –  Anon Jan 29 '13 at 0:04
    
Yes, we have to take into account $DLLL$, $DLDLL$, $DLLDL$, and so on. (There are $10$ of these like $3$, dislike the others.) And they are used in calculating the probability. –  André Nicolas Jan 29 '13 at 0:09
    
Your confusion comes from the following: You are calculating the event that he will like the first, second, and the third food, and then you say, "I don't care about the last two foods," and you put $2$ and $2$. Here (in your question), the order is not important. –  Anon Jan 29 '13 at 0:19
    
Because of this reason, your current solution does not take into account the case e.g. LDLLL, as you have mentioned. –  Anon Jan 29 '13 at 0:21

2 Answers 2

This to me looks like a Bernoulli trial with $p=1/2$.

Probability that your friend like $k=3$ of $n=5$ foods he tries is

$$P(n,k) = \binom{n}{k} p^k (1-p)^{n-k} \implies P(5,3) = \binom{5}{3} \left ( \frac{1}{2} \right )^3 \left ( \frac{1}{2} \right )^2 = \frac{5}{16}$$

Probability that he likes at least $3$ of $5$ foods he tries is

$$P(5,3) + P(5,4) + P(5,5) = \frac{10+5+1}{32} = \frac{1}{2}$$

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Suppose that he likes exactly $3$: there are $\binom53$ different sets of $3$ that he could like, and each of them occurs with probability $\left(\frac12\right)^5$, so the probability that he likes exactly $3$ is $\binom53\left(\frac12\right)^5$.

If he likes exactly $4$, there are $\binom54$ different sets of $4$ that he could like, and again each occurs with probability $\left(\frac12\right)^5$, so the probability that he likes exactly $4$ is $\binom54\left(\frac12\right)^5$.

There is only one way for him to like all $5$, and that again occurs with probability $\left(\frac12\right)^5$.

Thus, the probability that he likes at least $3$ is

$$\left(\binom53+\binom54+1\right)\left(\frac12\right)^5=\frac{16}{32}=\frac12\;.$$

This is a general approach that will work in a wide variety of settings. In this particular problem, however, there is a shortcup. Each sequence of likes and dislikes can be paired with its opposite, obtained by turning each $L$ into a $D$ and vice versa. Thus, $LLDLD$, for instance, is paired with $DDLDL$. Each pair contains one sequence with $3$ or more likes and one with $2$ or fewer, so exactly half of all sequences have $3$ or more likes, and the probability of getting one of these is therefore $\frac12$, since all sequences are equally likely.

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