One bag contains 4 white balls and 3 black balls. If someone draws one ball from the bag and then replaces it and draws another, what is the probability that both will be black?
A: 8/12
B: 9/49
C: 1/6
D: 3/7
Do you think the balls remember what happened to them on the first draw when you do the second draw ?
What is the prob of drawing a black ball the first time?
Is it A?
Why do you think it is A? are you guessing?
What is the prob of getting a black in the first draw?
the prob is 3/4 i think
OHHH IS IT D??
no and no
There are 7 balls in the bag of which 3 are black
So the prob of drawing a black the first time is 3/7
So what will happen the second draw??
What is the answer to the first question I asked you?
Oh b: 9/49
You got it!!
To solve this problem, we need to determine the probability of drawing a black ball on the first draw, and then drawing another black ball on the second draw.
The probability of drawing a black ball on the first draw is given by the number of black balls divided by the total number of balls in the bag. In this case, there are 3 black balls and a total of 7 balls (4 white + 3 black). Therefore, the probability of drawing a black ball on the first draw is 3/7.
Since the ball is replaced after the first draw, the probability of drawing a black ball on the second draw is also 3/7.
To calculate the probability of both events happening, we multiply the probabilities of each event together. Therefore, the probability of both balls being black is (3/7) * (3/7) = 9/49.
So the correct answer is B: 9/49.