A bag contains 8 white and 3 black counters. Two counters are drawn, one after the other. Find the probability of drawing one white and one black counter, in any order. i) If the first counter is replaced ii) if the first counter is not replaced
To find the probability of drawing one white and one black counter, in any order, we can calculate the probabilities separately for each case.
i) If the first counter is replaced:
When the first counter is replaced, it means that after each draw, the counter is put back into the bag before the next draw. In this case, the probability of drawing one white and one black counter will be the same, regardless of the order.
The probability of drawing a white counter on the first draw is 8/11, because there are 8 white counters out of 11 total counters in the bag.
After the first counter is drawn and replaced, there are still 11 counters in the bag, but now only 3 black counters. So the probability of drawing a black counter on the second draw is 3/11.
Since the order does not matter (i.e., white first and black second, or black first and white second), we can add the probabilities of the two possible outcomes: (8/11) * (3/11) + (3/11) * (8/11) = 24/121 + 24/121 = 48/121.
Therefore, the probability of drawing one white and one black counter, in any order, with replacement, is 48/121.
ii) If the first counter is not replaced:
When the first counter is not replaced, it means that after each draw, the counter is not put back into the bag before the next draw. In this case, the probability of drawing one white and one black counter will depend on the order of the draws.
The probability of drawing a white counter on the first draw is 8/11, as mentioned earlier.
After the first counter is drawn and not replaced, there are now only 10 counters in the bag, out of which 3 are black. So the probability of drawing a black counter on the second draw, given that the first counter was white, is 3/10.
Similarly, if we draw a black counter first, the probability of drawing a white counter on the second draw, given that the first counter was black, is 8/10.
Since the order matters (white first and black second, or black first and white second), we need to calculate the probabilities for both cases and then sum them up: (8/11) * (3/10) + (3/11) * (8/10) = 24/110 + 24/110 = 48/110.
Therefore, the probability of drawing one white and one black counter, in any order, without replacement, is 48/110, which simplifies to 24/55.