Answer:
[tex]\displaystyle P(A)=\frac{43}{90}\approx 0.48[/tex]
Step-by-step explanation:
Probability
There are 4 blue chips (B) and 4 red chips (R) in a hat. The event to analyze consists in two drawings from the hat, and it can be done in 4 different ways, as shown in the sample space below:
[tex]\Omega=\{BB,BR,RB,RR\}[/tex]
We are required to find the probability that the second chip is blue. There are two favourable results to form the event as follows
[tex]A=\{BB,RB\}[/tex]
Let's analyze the first option BB. There are 4 blue chips out of 8 in total, thus the probability to draw a blue chip is
[tex]\displaystyle \frac{4}{8}=\frac{1}{2}[/tex]
Once extracted a blue chip, we replace it and add another blue chip, thus the new contents of the hat is: 4 red chips and 5 blue chips. Now we find the probability to draw a second blue chip (5 out of 9):
[tex]\displaystyle \frac{5}{9}[/tex]
The combined probability of both events is
[tex]\displaystyle P(BB)=\frac{1}{2}\cdot \frac{5}{9}=\frac{5}{18}[/tex]
Now for the second option RB. The probability to first draw a red chip is
[tex]\displaystyle \frac{4}{8}=\frac{1}{2}[/tex]
Once we picked a red chip, we replace it along with two new red chips, so the new contents of the hat is: 4 blue chips and 6 red chips. The probability to draw a blue chip in these conditions is
[tex]\displaystyle \frac{4}{10}=\frac{2}{5}[/tex]
The combined probability of both events is
[tex]\displaystyle P(RB)=\frac{1}{2}\cdot \frac{2}{5}=\frac{1}{5}[/tex]
The required probability is the sum of both separate options
[tex]\displaystyle P(A)=\frac{5}{18}+\frac{1}{5}=\frac{43}{90}[/tex]
[tex]\boxed{\displaystyle P(A)=\frac{43}{90}\approx 0.48}[/tex]
