a fair coin is tossed 5 times . what is the probability that it lands heads up at least twice ?
fair coin is tossed 5 times . hence total number of outcomes = 2 ^ 5 = 32 . problem asks for probability of getting atleast heads twice . hence if we calculate probability of getting heads exactly once and probability of not getting heads at all and subract it from the total probability of the event which is 1 ( as total probability of certain event will be always 1 ) we can get the probability of getting atleast heads twice . probability of getting exactly one head and no heads = ( number of possible outcomes [ htttt , thttt , tthtt , tttht , tttth , ttttt ] = 6 ) / ( total possible outcomes = 32 ) = > 6 / 32 = 3 / 16 hence probability of getting atleast heads twice = 1 - ( 3 / 16 ) = 13 / 16 = > choice [ d ]
divide(add(add(add(choose(5, const_2), choose(5, const_3)), choose(5, const_4)), choose(5, 5)), power(const_2, 5))