Consider two events #A# and #B#, where #A# is the event that first bean taken from the jar is cherry and #B# is the event that second bean taken from the jar is cherry. What is the probability of #P(AnnB)# i.e. both are cherries. Well the formula is
#P(AnnB)=P(A)xxP(B|A)#, where #P(A)# is obviously #14/38# as there are #14# cherry jelly beans among a total of #38# jelly beans. #P(B|A)# is the probability of getting a cherry jelly bean, subject to the condition that #A# event has already happened. If it is so we now have #13# cherry jelly beans out of #37# and hence #P(B|A)=13/37# and
#P(AnnB)=14/38xx13/37#
Similarly for four cherry jelly beans,
the probability of getting first cherry jelly bean is #14/38#
the probability of getting second cherry jelly bean would be #13/37#
the probability of getting third cherry jelly bean would be #12/36#
and the probability of getting fourth cherry jelly bean would be #11/35#
Hence the probability of grabbing #4# beans from the jar at the same time would be
#14/38xx13/37xx12/36xx11/35=7/19xx13/37xx1/3xx11/35=1001/73815=0.0136# up to #4dp#
