Prc1995 wrote:
DMilwaukee wrote:
Probability of making at-least one is (1-(probability of making none))
probability that 1st and 2nd shot is missed = 3/4 * 3/4 = 9/16
1-9/16 = 7/16
What about the no. of shoots she gets ? the statement says that she misses the first shot and their is a probability of 50% on that. And their are only 3 shots she gets, out of which one is a fail so there has to be one success from the next two ?
let " not scored" be designated as NS
let " scored" be designated as S
we have 3 throws,
Possible outcomes are as follows
1.NS ,S, S = ( 1/2 , 1/4 ,3/4)
2.NS ,NS, S = ( 1/2 ,3/4 ,1/4)
3.NS , S , NS = (1/2 ,1/4 ,1/4)
however, we are only interested in the 2 subsequent throws
therefore,probability of the two subsequent throws = 1/4 *3/4 +3/4 *1/4 +1/4 *1/4 = 7/16
hence, ans= 7/16