Question

in a bag there are 4 red and 3 black balls. what is the probability of drawing the first ball and second black, third red and fourth black and so on, if they are drawn one at a time​

Answer 1

Answer:

Step-by-step explanation:

This is a conditional probability. I will define A as the first ball being black and B as the second ball being red. Therefore, we want this:

P(A∣B)  

Using the definition of conditional probability, that becomes:

P(A∣B)=P(A∩B)P(B)  

Using the law of total probability, that becomes:

P(A∣B)=P(A∩B)P(A∩B)+P(A∁∩B)  

Now we need those 2 probabilities: From my definitions at the start,  P(A∩B)  is the probability of having black then red. The probability of the first being black is 3/7. After that, there will be 2 black and 4 red remaining, so the probability of drawing a red after a black is 4/6. Therefore this probability is the product of the 2, which is 2/7.

P(A∁∩B)  is the probability that both are red. The probability that the first is red is 4/7. After that, there will be 3 red and 3 black remaining, so the probability of the second being red when the first is red is 3/6. The probability of both being red is the product of those, 2/7. Now that we have those, we can calculate the answer.

P(A∣B)=P(A∩B)P(A∩B)+P(A∁∩B)  

P(A∣B)=2/72/7+2/7=1/2