Question

According to an article, 19.6% of Internet stocks that entered the market in 1999 ended up trading below their initial offering prices. If you were an investor who purchased three Internet stocks at their initial offering prices, what was the probability that at least two of them would end up trading at or above their initial offering price? (Round your answer to four decimal places.) 

Answer 1

Solution :

Given that ,

p = 0.196

1 - p = 1 - 0.196 = 0.804

n = 3

p (  x $\geq$ 2 )  =p (x = 2 )+ p (x = 3)

Using binomial probability formula ,

P(X $\geq$ 2) = ((n! / (n - x)!) * p^x * (1 - p)^{n - x}

P(X = 2) = (3! / (3 - 2)!) * 0.196² * 0.196)^{3 - 2}

p (  x = 2 )  = 0.9266

Probability = 0.9266

P(X = 3) = (3! / (3 - 3)!) * 0.1963 * 0.196)3 - 3

p (  x = 3 )  = 0.7530

Probability = 0.7530

= 0.0927 + 0.0075

p (  x $\geq$ 2 )  = 0.1002

Probability = 0.1002