Suppose I have 8 coins.4 of them have heads on both sides and the other 4 have tails on both sides. I reach in and get 4 coins at one time. What's the probability that I get 2 head coins and 2 tails?
Well I was thinking that it would be 1/2 since both heads and tails have 4 coins each.
To find the probability of drawing 2 head coins and 2 tail coins when selecting 4 out of the 8 given coins, we can use the concept of combinations.
First, let's determine the total number of possible ways to choose 4 coins out of 8. This can be calculated using the formula for combinations, denoted as "nCr", where n is the total number of objects and r is the number of objects being chosen.
In this case, we have 8 coins and we want to choose 4 of them, so the total number of ways to select 4 coins out of 8 is:
8C4 = (8!)/(4!(8-4)!) = 70
Now, let's determine the number of ways to choose 2 head coins and 2 tail coins. We have 4 head coins and 4 tail coins, so we can choose 2 of each, which can be calculated as:
4C2 * 4C2 = (4!)/(2!(4-2)!) * (4!)/(2!(4-2)!) = 6 * 6 = 36
Finally, to find the probability, we divide the number of desired outcomes (choosing 2 head and 2 tail coins) by the total number of possible outcomes (choosing any 4 coins):
Probability = Number of desired outcomes / Total number of possible outcomes
Probability = 36/70
Probability ≈ 0.5143
Therefore, the probability of selecting 2 head coins and 2 tail coins when choosing 4 coins is approximately 0.5143 or 51.43%.