A coin is tossed $10$ times. The probability of getting exactly six heads is
- $\dfrac{{512}}{{513}}$
- $\dfrac{{105}}{{512}}$
- $\dfrac{{100}}{{153}}$
- $^{10}{C_8}$
The probability of getting exactly k successes in n trials is given by the binomial formula: C(n, k) * p^k * (1-p)^(n-k). Here, n=10, k=6, and p=0.5. Calculating this gives C(10, 6) * (0.5)^10 = 210 / 1024, which simplifies to 105/512.
Using the formula for binomial probability, the probability of getting exactly r successes in n trials is given by nCr multiplied by p to the power of r multiplied by q to the power of n minus r. For a fair coin, the probability of heads p is one half and tails q is one half. Substituting n equals 10 and r equals 6 gives the probability as 10C6 times one half to the power of 6 times one half to the power of 4. Evaluating this gives 210 multiplied by one over 1024, which simplifies to 105 over 512.