probability

Question

if a bag contains 342 blue marbles, and 121 red ones, what is the prob of picking etiher a blue r red marble in one turn?

Answer 1

Probability of an event (E) can be restated as a ratio...
"(# of favorable outcomes) / (# of total outcomes)", or....

P(E)=n(E)/n(S)........Eq1, where...

P(E) is the probability of the event E
n(E) is the # of favorable outcomes of an event E
n(S) is the # of total outcomes in set S

Let...
n(B)=blue=342
n(R)=red=121
n(S)=n(B)+n(R)
      =342+121......substitute values
      =463.............total # of marbles in the set

P(B)=n(B)/n(S)
      =342/463.......can reduce fraction to get B:S "chance" ratio
      =0.739
      =73.9%.........chance of picking a blue marble with 1st pick

P(R)=n(R)/n(S)
      =121/463.......can reduce fraction to get R:S "chance" ratio
      =0.261
      =26.1%.........chance of picking a red marble with 1st pick

P(B or R)=P(B) + P(R)
             =73.9% + 26.1%
             =100%...chance of picking either a red OR blue marble

                           1st pick, and this makes sense as there are

                           only blue and red marbles to pick from.

Note that subsequent chances of picking colors depends on whether
the previous pick is returned to the bag. If not, then the values
listed above [n(S), n(B), n(R)] need to be adjusted and, using Eq1
and the updated values, the percentages will change.

Answer 2

Probability of an event (E) can be restated as a ratio...

"(# of favorable outcomes) / (# of total outcomes)", or....

 

P(E)=n(E)/n(S)........Eq1, where...

 

P(E) is the probability of the event E

n(E) is the # of favorable outcomes of an event E

n(S) is the # of total outcomes in set S

 

Let...

n(B)=blue=342 

n(R)=red=121

n(S)=n(B)+n(R)

      =342+121......substitute values

      =463.............total # of marbles in the set

 

P(B)=n(B)/n(S)

      =342/463.......can reduce fraction to get B:S "chance" ratio

      =0.739

      =73.9%.........chance of picking a blue marble with 1st pick

 

P(R)=n(R)/n(S)

      =121/463.......can reduce fraction to get R:S "chance" ratio

      =0.261

      =26.1%.........chance of picking a red marble with 1st pick

 

P(B or R)=P(B) + P(R)

             =73.9% + 26.1%

             =100%...chance of picking either a red OR blue marble 

                           1st pick, and this makes sense as there are only 

                           blue and red marbles to pick from.

 

Note that subsequent chances of picking colors depends on whether 

the previous pick is returned to the bag.  If not, then the values 

listed above [n(S), n(B), n(R)] need to be adjusted and, using Eq1

and the updated values, the percentages will change.