Kevin flips a coin four times. what is the probability that he gets heads on at least one of the four flip?
a) \(\frac{1}{16}\)
b) \(\frac{1}{4}\)
c) \(\frac{3}{4}\)
d) \(\frac{13}{16}\)
e) \(\frac{15}{16.}\)
books says its option 'e' but i say its 'b'
what do you think?
justifiction:
[p(H)*p(T)*p(T)*p(T)] + [p(H)*p(H)*p(T)*p(T)] + [p(H)*p(H)*p(H)*p(T)] + [p(H)*p(H)*p(H)*p(H)]
=> [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] = 4/16 => 1/4.
thank you
siva