1

I keep on getting different answers for this problem. Can someone help?

A fair coin is tossed 100 times. What is the probability that more than half of them are heads?

$ P(X \geq 51) = 1 - P(X < 51) = 1 - binomcdf(100, 0.50, 50) = 0.460$

But when I use $ P(X \geq 51) = 1 - P(X < 51) = 1 - normalcdf(50.5, e99, 50,5) = 0.5398$

I used $np = 100\cdot 0.50 = 50$ and $\sqrt{npq} = 5$

EDIT:

I got 0.460 for both now. $ P(X \geq 51) = 1 - P(X < 51) = 1 - normalcdf(-e99, 50.5, 50,5) = 0.460$

Guest
  • 167

1 Answers1

1

The probability that exactly $50$ of the coins are heads is ${100 \choose 50} / \sum_{j=0}^{100} {100 \choose j}$. If you subtract this from one you get the probability that the majority of the coins are either heads are tails. Since heads and tails are equally likely, you divide by 2 to get the probability that 51 or more coins are heads. The answer should certainly be less than $1/2$, because you subtract from 1 and then divide by 2 to get the answer.

user2566092
  • 26,142