In a class of TOTAL,
there are SPECIAL students who
EVENT_PY.
If the teacher chooses CHOSEN
students, what is the probability that
ALL ? ((CHOSEN === 2) ? "both" : "all " + cardinalThrough20(CHOSEN)) :
((CHOSEN === 2) ? "neither" : "none of the " + cardinalThrough20(CHOSEN))
of them EVENT_PY?
We can think about this problem as the probability of
CHOSEN events happening.
The first event is the teacher choosing one student
who EVENT_S. The second event is the teacher
choosing another student who EVENT_S, given
that the teacher already chose someone who
EVENT_S
, and so on.
We can think about this problem as the probability of
CHOSEN events happening.
The first event is the teacher choosing one student
who EVENT_S. The second event is the teacher
choosing another student who EVENT_S, given
that the teacher already chose someone who
EVENT_S
.
The probabilty that the teacher will choose someone
who EVENT_S is the number of students who
EVENT_P divided by the total number of
students: \dfrac{PROBS[0][0]}
{PROBS[0][1]}.
Once the teacher's chosen one student, there are only
TOTAL-1 left.
There's also one fewer student who EVENT_S, since the teacher isn't going to pick the same student twice.
So, the probability that the teacher picks a second
student who also EVENT_S is
\dfrac{PROBS[1][0]}
{PROBS[1][1]}.
The probability of the teacher picking two
students who EVENT_P must then be
\dfrac{PROBS[0][0]}
{PROBS[0][1]} \cdot
\dfrac{PROBS[1][0]}
{PROBS[1][1]}.
We can continue using the same logic for the rest of the students the teacher picks.
So, the probability of the teacher picking
CHOSEN students such that
ALL ? ((CHOSEN === 2) ? "both" : "all")
: "none" of them
EVENT_PY is:
\qquad_.map(PROBS, function(p){
return "\\dfrac{"+p[0]+"}{"+p[1]+"}";
}).join("\\cdot") =
\dfrac{ANS_N}{ANS_D}
= fractionReduce(ANS_N, ANS_D)
A CONTAINER contains
REDMAR red MARBLEs,
GREENMAR green MARBLEs,
and BLUEMAR
blue MARBLEs.
If we choose a MARBLE, then another MARBLE without putting the first one back in the CONTAINER, what is the probability that the first MARBLE will be COLOR_ONE and the second will be COLOR_TCOLOR_ONE === COLOR_T ? " as well?" : "?"
The probability of event A happening, then event B, is
the probability of event A happening times
the probability of event B happening given that
event A already happened.
In this case, event A is picking a COLOR_ONE
MARBLE and leaving it out.
Event B is picking COLOR_TWO
MARBLE.
Let's take the events one at at time. What is the probability that the first MARBLE chosen will be COLOR_ONE?
There are NUM_ONE
COLOR_ONE MARBLEs,
and TOTAL total, so the
probability we will pick a COLOR_ONE
MARBLE is
\dfrac{NUM_ONE}
{TOTAL}.
After we take out the first MARBLE, we
don't put it back in, so there are only
TOTAL-1 MARBLEs
left.
Also, we've taken out one of the COLOR_ONE
MARBLEs, so there are only
AFTER_NUM left altogether.
Since the first MARBLE was
COLOR_ONE, there are still
AFTER_NUM
COLOR_T MARBLEs
left.
So, the probability of picking COLOR_TWO
MARBLE after taking out a
COLOR_ONE MARBLE is
\dfrac{AFTER_NUM}
{TOTAL-1}.
Therefore, the probability of picking a
COLOR_ONE MARBLE, then
COLOR_TWO MARBLE
is \dfrac{NUM_ONE}{TOTAL}
\cdot \dfrac{AFTER_NUM}{TOTAL-1}
= \dfrac{ANS_N}{ANS_D}
= fractionReduce(ANS_N, ANS_D)
Captain person(1) has a ship, the H.M.S Crimson Lynx. The ship is DISTANCE furlongs from the dread pirate person(2) and his merciless band of thieves. The ship is DISTANCE furlongs from the dread pirate person(2) and her merciless band of thieves.
If his ship hasn't already been hit, Captain person(1) has probability
C_HIT_PRETTY of hitting the pirate ship.
If his ship has been hit, Captain person(1) will always miss.
If her ship hasn't already been hit, Captain person(1) has probability
C_HIT_PRETTY of hitting the pirate ship.
If her ship has been hit, Captain person(1) will always miss.
If his ship hasn't already been hit, dread pirate person(2) has probability
P_HIT_PRETTY of hitting the Captain's ship.
If his ship has been hit, dread pirate person(2) will always miss.
If her ship hasn't already been hit, dread pirate person(2) has probability
P_HIT_PRETTY of hitting the Captain's ship.
If her ship has been hit, dread pirate person(2) will always miss.
If the Captain and the pirate each shoot once, and the Captain shoots first, what is the probability that QUESTION?
If the Captain and the pirate each shoot once, and the pirate shoots first, what is the probability that QUESTION?
The probability of event A happening, then event B, is the probability of
event A happening times the probability of event B happening
given that event A already happened.
In this case, event A is EV_A and event B is EV_B.
The Captain fires first, so his ship can't be sunk before he fires his cannons. The pirate fires first, so his ship can't be sunk before he fires his cannons.
The Captain fires first, so her ship can't be sunk before she fires her cannons. The pirate fires first, so her ship can't be sunk before she fires her cannons.
So, the probability of EV_A is PR_A.
If the Captain hit the pirate ship, the pirate has no chance of firing back. If the pirate hit the Captain's ship, the Captain has no chance of firing back.
If the Captain missed the pirate ship, the pirate has a normal chance to fire back. If the pirate missed the Captain's ship, the Captain has a normal chance to fire back.
So, the probability of EV_B given EV_A is PR_B.
The probability that QUESTION is then the probability of EV_A times the probability of EV_B given EV_A.
This is PR_A \cdot PR_B
= fraction(ANS_N, ANS_D, true, true)