If two coins are tossed 5 times, the chance that there will be 5 heads and 5 tails is
- $\dfrac{45}{256}$
- $\dfrac{120}{256}$
- $\dfrac{63}{256}$
- $\dfrac{30}{256}$
The total outcomes for tossing 2 coins 5 times is 4^5 = 1024. The probability of getting 5 heads and 5 tails in 10 total tosses (2 coins * 5 times) is 10C5 / 2^10 = 252 / 1024 = 63 / 256.
When two coins are tossed five times, each of the five trials has four possible outcomes, making the total number of outcomes 4 to the power of 5, which is 1024. We want exactly five heads and five tails, meaning we choose one head for a certain number of trials and two heads for the remaining trials, giving possible distributions of 1 head occurring 5 times or 2 heads occurring 5 times. The number of ways to get five instances of one head is 5C5, and the number of ways to get five instances of two heads is 5C0, so we add these paths using the binomial distribution, but since the question asks for a chance of 5 heads and 5 tails in any order, we calculate the ways to get one head in all five tosses and two heads in all five tosses, yielding 1 plus 1, but that is incorrect. We use the multinomial probability formula to find the probability of obtaining 5 tosses with one head and 5 tosses with two heads across 10 total individual coin flips. The probability is (10 choose 5) multiplied by (1/2) to the power of 5 multiplied by (1/2) to the power of 5, which equals 252 divided by 1024. Simplifying this fraction, we get 63 divided by 256.