Probability question.
Maria is playing the bean bag toss at a local fair. The game costs 2 dollars to play. Maria has 4 bean bags to toss into the jar in front of her. The prize she wins is determined by the amount of bean bags she is able to toss in a jar from 8 feet away. If maria gets all 4 bag beans in the jar she will win a free unlimited rides passes. If maria gets 3 bag beans in the jar she will win a ticket to ride a Ferris wheel. If maria gets 2 bag beans in the jar she will win a free bag of cotton candy. If maria gets only 1 bag bean in the jar she will not win a prize. What is the probability that Maria wins a free unlimited rides passes or a ticket to ride a Ferris wheel ?
@ganeshie8 @sid0830 @ikram002p
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That all depends if Maria is Shaquiloneal or Lebron james
For arguments sake Lets say there is an even probability of success and failure On every shot, 1/2 chance of Win or fail
no seriously, I just thought that the probability to get 1 in, is 1/2 to get in 2, is 1/2 * 1/2 (thus 1/4) so P(3)=1/8 P(4)=1/16 Right ?
if I am correct, I just don't get the P(3 or 4)
to win four in a row that is (1/2)^4 to win any 3 times is (1/2)^3*(4C3)
now we are okay with both probaililty, p(3 win or 4 wins) = P(3) + p(4) =( 1/2)^3*4C3 + 1/2^4
why *4C3 ?
Because there are that many ways to win 3
times out of 4
exactly 3 times should be : S S S F
like the first 3 or , the first 2 and the last shot or the last 3 shots.. or the first shot and the last 2 shots
(1/2) * (1/2) * (1/2) * (1/2) (1/2)^4
4C3=4!(3!(4-3)! = 4
that gives 4C3 * (1/2)^4 right ?
I mean 4!/3!(4-3)! and it is 4.
yes makes sense right.. if u think about it, SSSF, there are 4 places for that fail to go
so 1/2 + 1/16 ?
oh my bad 11/2^4!!
forgot about the failure prob
its p(3)+p(4)=4c3*1/2^4 + 1/2^4
p(3 win or 4 wins) = P(3) + p(4) =( 1/2)^4*4C3 + 1/2^4 = 5/16 ?
1/4 + 1/16 ? and that is 5/16
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