Answer:
E[X] = (np) (p+(1-p))^(n-1)
With n =10,000 and p=0.50 we get
E[X]= (10,000*0.50) (0.50+(1–0.50))^(10,000–1)
E[X] = 5,000(1)
So assuming the coin is fair (p=50%), then we can expect to get heads 5,000 times when the coin is tossed 10,000 times.
Step-by-step explanation:
