If three unbiased coins are tossed, find the probability ...
If three unbiased coins are tossed, find the probability that they are all heads
\(\frac{1}{2}\)
\(\frac{1}{3}\)
\(\frac{1}{9}\)
\(\frac{1}{8}\)
Correct answer is D
P(H) = \(\frac{1}{2}\) and P(T) = \(\frac{1}{2}\)
Using the binomial prob. distribution,
(H + T)3 = H3 + 3H2T1 + 3HT2 + T3
Hence the probability that three heads show in a toss of the three coins is H3
= (\(\frac{1}{2}\))3
= \(\frac{1}{8}\)
Convert 3.1415926 to 5 decimal places ...
Evaluate \(\frac{0.00000231}{0.007}\) and leave the answer in standard form...
If y = (1 - 2x)\(^3\), find the value of dy/dx at x = -1 ...
P = {x:(x - 2)(x + 3) (x - 4)(x + 5) = 0}. The elements in P are ...
Solve for x and y in the equations below x2 - y2 = 4 x + y = 2...