A set of 10 pills contains 5 placebo pills. (i) 5 pills are randomly selected from the set. What is the probability that at least 2 of them are placebo pills? (ii) Pills are drawn one at a time without replacement. What is the probability that the first three selected pills are all placebo pills?

1. Problem Description

A set of 10 pills contains 5 placebo pills.

(i) 5 pills are randomly selected from the set. What is the probability that at least 2 of them are placebo pills?

(ii) Pills are drawn one at a time without replacement. What is the probability that the first three selected pills are all placebo pills?

2. Solution Steps

(i) Let X be the number of placebo pills in the sample of 5 pills. We want to find $P(X \geq 2)$. It is easier to calculate the complement: $P(X \geq 2) = 1 - P(X < 2) = 1 - [P(X=0) + P(X=1)]$.

The total number of ways to select 5 pills out of 10 is $\binom{10}{5}$.

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

$P(X=0)$ is the probability that none of the selected pills are placebo pills. This means we selected 5 pills from the 5 non-placebo pills.

$P(X=0) = \frac{\binom{5}{5}}{\binom{10}{5}} = \frac{1}{252}$.

$P(X=1)$ is the probability that exactly 1 pill is a placebo pill. This means we selected 1 placebo pill from the 5 placebo pills and 4 non-placebo pills from the 5 non-placebo pills.

$P(X=1) = \frac{\binom{5}{1} \binom{5}{4}}{\binom{10}{5}} = \frac{5 \cdot 5}{252} = \frac{25}{252}$.

$P(X \geq 2) = 1 - [P(X=0) + P(X=1)] = 1 - \left[\frac{1}{252} + \frac{25}{252}\right] = 1 - \frac{26}{252} = \frac{252-26}{252} = \frac{226}{252} = \frac{113}{126}$.

(ii) We want to find the probability that the first three pills selected are all placebo pills.

The probability that the first pill is a placebo pill is $\frac{5}{10} = \frac{1}{2}$.

Given that the first pill is a placebo pill, there are now 4 placebo pills and 9 total pills remaining.

The probability that the second pill is a placebo pill, given that the first pill was a placebo pill, is $\frac{4}{9}$.

Given that the first two pills were placebo pills, there are now 3 placebo pills and 8 total pills remaining.

The probability that the third pill is a placebo pill, given that the first two pills were placebo pills, is $\frac{3}{8}$.

The probability that the first three pills selected are all placebo pills is:

$\frac{5}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} = \frac{1}{2} \cdot \frac{4}{9} \cdot \frac{3}{8} = \frac{12}{144} = \frac{1}{12}$.

3. Final Answer

(i) $\frac{113}{126}$

(ii) $\frac{1}{12}$