In a class of

,
there are `TOTAL`

students who
`SPECIAL``EVENT_PY`.

If the teacher chooses

students, what is the probability that
`CHOSEN``ALL ? ((CHOSEN === 2) ? "both" : "all") :
((CHOSEN === 2) ? "neither" : "none")`
of them `EVENT_PY`?

We can think about this problem as the probability of

events happening.
`CHOSEN`

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`
`CHOSEN > 2 ? ", and so on." : "."`

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

left.
`TOTAL-1`

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

students such that
`CHOSEN``ALL ? ((CHOSEN === 2) ? "both" : "all")
: "none"` of them
`EVENT_PY` is
`_.map(PROBS, function(p){
return "\\dfrac{"+p[0]+"}{"+p[1]+"}";
}).join("\\cdot")` =
\dfrac{`ANS_N`}{`ANS_D`}

A `CONTAINER` contains

red `REDMAR``MARBLE`s,

green `GREENMAR``MARBLE`s,
and

blue `BLUEMAR``MARBLE`s.

If we choose a `MARBLE`, then another `
MARBLE`
*without putting the first one back in the
CONTAINER*, what is the probability that
the first

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` `MARBLE`s,
and

total, so the
probability we will pick a `TOTAL``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``MARBLE`s
left.

Also, we've taken out one of the `COLOR_ONE`
`MARBLE`s, so there are only

left altogether.
`AFTER_NUM`

Since the first `MARBLE` was
`COLOR_ONE`, there are still
`AFTER_NUM``COLOR_T` `MARBLE`s
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`}

Captain `person(1)` and `his(1)` ship, the H.M.S Crimson Lynx, are two furlongs
from the dread pirate `person(2)` and `his(2)` merciless band of scallawags.

The Captain has probability

of hitting the pirate ship, if `C_HIT_PRETTY``his(1)`
ship hasn't already been hit. If it has been hit, `he(1)` will always miss.
The pirate has probability

of hitting the
Captain's ship, if `P_HIT_PRETTY``his(2)` ship hasn't already been hit. If it has been hit, `he(2)` will always
miss as well.

If the `C_FIRST ? "Captain" : "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 `C_FIRST ? "Captain" : "pirate"` fires first, so
`his(C_FIRST ? 1 : 2)` ship can't be sunk before
`he(C_FIRST ? 1 : 2)` fires his cannons.

So, the probability of `EV_A` is

.
`PR_A`

If the `C_FIRST ? "Captain" : "pirate"` hit the `C_FIRST ? "pirate" : "Captain's"` ship,
the `C_FIRST ? "pirate" : "Captain"` has no chance of firing back.

If the `C_FIRST ? "Captain" : "pirate"`
missed the `C_FIRST ? "pirate" : "Captain's"`
ship,
the `C_FIRST ? "pirate" : "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)`