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 they are not black?

Explanation:

Step1: Find number of non - black socks

First, we calculate the number of non - black socks. The total number of socks is 40, and the number of black socks is 20. So the number of non - black socks is \(16 + 4=20\).

Step2: Calculate probability of first non - black sock

The probability of drawing a non - black sock on the first draw is the number of non - black socks divided by the total number of socks. So \(P(\text{1st non - black})=\frac{20}{40}\).

Step3: Calculate probability of second non - black sock (without replacement)

After drawing one non - black sock, the number of non - black socks left is \(20 - 1 = 19\) and the total number of socks left is \(40-1=39\). So the probability of drawing a non - black sock on the second draw given that the first was non - black is \(P(\text{2nd non - black}|\text{1st non - black})=\frac{19}{39}\).

Step4: Calculate the probability of two non - black socks

Since we want the probability that both socks are non - black (and we are using the multiplication rule for dependent events \(P(A\cap B)=P(A)\times P(B|A)\)), the probability that both socks are non - black is \(P=\frac{20}{40}\times\frac{19}{39}\)

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We can also think of it as the number of ways to choose 2 non - black socks divided by the number of ways to choose 2 socks from 40. The number of ways to choose 2 non - black socks is \(C(20,2)=\frac{20!}{2!(20 - 2)!}=\frac{20\times19}{2\times1}\), and the number of ways to choose 2 socks from 40 is \(C(40,2)=\frac{40!}{2!(40 - 2)!}=\frac{40\times39}{2\times1}\). Then \(P=\frac{C(20,2)}{C(40,2)}=\frac{\frac{20\times19}{2}}{\frac{40\times39}{2}}=\frac{20\times19}{40\times39}=\frac{19}{78}\)

Answer:

\(\frac{19}{78}\)