Question

fred reaches into a bag of 40 jumbled pairs of socks, in which there are 20 black socks, 16 brown ones and 4 blue ones. if he takes two socks without replacement, what is the probability that one is black and the other not black? 1st sock 2nd sock black not black black not black

Explanation:

Step1: Calculate probability of Black then Not Black

The probability of first sock being black is $\frac{20}{40}$. After taking a black sock, there are 39 socks left, and 20 non - black socks. So the probability of second sock being non - black is $\frac{20}{39}$. The probability of this path is $\frac{20}{40}\times\frac{20}{39}$.

Step2: Calculate probability of Not Black then Black

The probability of first sock being non - black is $\frac{20}{40}$ (since total non - black socks are $16 + 4=20$). After taking a non - black sock, there are 39 socks left, and 20 black socks. So the probability of second sock being black is $\frac{20}{39}$. The probability of this path is $\frac{20}{40}\times\frac{20}{39}$.

Step3: Sum the two probabilities

We add the probabilities of the two mutually exclusive paths (Black then Not Black and Not Black then Black). So total probability $P=\frac{20}{40}\times\frac{20}{39}+\frac{20}{40}\times\frac{20}{39}$.
Simplify the expression:
First, $\frac{20}{40}=\frac{1}{2}$. So $P = 2\times(\frac{1}{2}\times\frac{20}{39})=\frac{20}{39}$.

Answer:

$\frac{20}{39}$