`randFromArray(["bag", "jar", "box", "cup"])`
`randFromArray(["marble", "ball", "jelly bean"])`
`randRange(3, 11)`
`randRange(3, 11)`
`randRange(3, 11)`
`RED + GREEN + BLUE`
`rand(2) == 0`
`randFromArray([["red", RED], ["green", GREEN], ["blue", BLUE]])`
`NOT ? TOTAL - CHOSEN_NUMBER : CHOSEN_NUMBER`

A `CONTAINER` contains `RED`

red `MARBLE`s,
`GREEN`

green `MARBLE`s, and `BLUE`

blue `MARBLE`s.

If a `MARBLE` is randomly chosen, what is the probability
that it is *not* `CHOSEN_COLOR`? Write your answer as a simplified fraction.

`NUMBER / TOTAL`

There are `RED` + `GREEN` + `BLUE` = `TOTAL`

`MARBLE`s in the `CONTAINER`.

There are `CHOSEN_NUMBER`

`CHOSEN_COLOR` `MARBLE`s.
That means `TOTAL` - `CHOSEN_NUMBER` = `NUMBER`

are *not* `CHOSEN_COLOR`.

The probability is `\displaystyle ``fractionSimplification(NUMBER, TOTAL)`

.

`randFromArray([
["a 1", [1]],
["a 2", [2]],
["a 3", [3]],
["a 4", [4]],
["a 5", [5]],
["a 6", [6]],
["at least 2", [2, 3, 4, 5, 6]],
["at least 3", [3, 4, 5, 6]],
["at least 4", [4, 5, 6]],
["more than 2", [3, 4, 5, 6]],
["more than 3", [4, 5, 6]],
["more than 4", [5, 6]],
["less than 4", [1, 2, 3]],
["less than 5", [1, 2, 3, 4]],
["less than 6", [1, 2, 3, 4, 5]],
["even", [2, 4, 6]],
["even", [2, 4, 6]],
["odd", [1, 3, 5]],
["odd", [1, 3, 5]],
["prime", [2, 3, 5]]
])`
`RESULT_POSSIBLE.length`

A fair six-sided die is rolled. What is the probability that the
result is `RESULT_DESC`? Write your answer as a simplified fraction.

`RESULT_COUNT / 6`

When rolling a die, there are `6`

possibilities: 1, 2, 3, 4, 5, and 6.

In this case, only `1`

result is favorable: the number `RESULT_POSSIBLE[0]`.

In this case, `RESULT_COUNT`

results are favorable: `toSentence(RESULT_POSSIBLE)`.

The probability is `\displaystyle ``fractionSimplification(RESULT_COUNT, 6)`

.

`randFromArray([
[3, "no heads", [0]],
[3, "heads exactly once", [1]],
[3, "heads exactly twice", [2]],
[3, "heads at least once", [1, 2, 3]],
[3, "heads at least twice", [2, 3]],
[3, "heads every time", [3]],
[4, "no heads", [0]],
[4, "heads exactly once", [1]],
[4, "heads exactly twice", [2]],
[4, "exactly three heads", [3]],
[4, "heads at least once", [1, 2, 3, 4]],
[4, "heads at least twice", [2, 3, 4]],
[4, "at least three heads", [3, 4]],
[4, "heads every time", [4]],
[3, "no tails", [3]],
[3, "tails exactly once", [2]],
[3, "tails exactly twice", [1]],
[3, "tails at least once", [0, 1, 2]],
[3, "tails at least twice", [0, 1]],
[3, "tails every time", [0]],
[4, "no tails", [4]],
[4, "tails exactly once", [3]],
[4, "tails exactly twice", [2]],
[4, "exactly three tails", [1]],
[4, "tails at least once", [0, 1, 2, 3]],
[4, "tails at least twice", [0, 1, 2]],
[4, "at least three tails", [0, 1]],
[4, "tails every time", [0]]
])`
`coinFlips(REPS)`
`(function() {
return jQuery.map(ALL, function( el, i ) {
return el[0];
});
})()`
`(function() {
return jQuery.map(jQuery.grep(ALL, function( el, i ) {
return WANTED.indexOf(el[1]) !== -1;
}), function( el, i ) {
return el[0];
});
})()`
`choose(REPS, WANTED)`
`pow(2, REPS)`

A fair coin is flipped `REPS == 3 ? "three" : "four"` times. What is the
probability of getting `DESC`? Write your answer as a simplified fraction.

`WANTED_COUNT / TWO_TO_REPS`

There are `(new Array(REPS)).join("2 \\cdot ")`2 = 2^{`REPS`} = `TWO_TO_REPS`

possibilities for the sequence of flips.

The possibilities are `toSentence(ALL_SEQS)`.

There `WANTED_COUNT == 1 ? "is only" : "are"` `plural(WANTED_COUNT, "favorable outcome")`: `toSentence(WANTED_LIST)`.

The probability is `\displaystyle ``fractionSimplification(WANTED_COUNT, TWO_TO_REPS)`

.

`randFromArray([ [1, 10], [11, 20], [21, 30], [31, 40], [41, 50], [51, 60], [61, 70], [71, 80], [81, 90], [91, 100] ])`
`(function() {
var list = [];
for (var i = LOW; i <= HIGH; i++) {
list.push(i);
}
return list;
})()`
`randFromArray([
["prime", KhanUtil.isPrime],
["divisible by both 2 and 3", function(n) { return n % 6 <= 0.5; }],
["divisible by either 3 or 5", function(n) { return n % 3 <= 0.5 || n % 5 <= 0.5; }],
["divisible by either 4 or 7", function(n) { return n % 4 <= 0.5 || n % 7 <= 0.5; }]
])`
`jQuery.grep(POSSIBLE, COND_FN)`
`WANTED_LIST.length`

A positive integer is picked randomly from `LOW` to `HIGH`, inclusive.

What is the probability that it is `COND_DESC`? Write your answer as a simplified fraction.

`WANTED_COUNT / POSSIBLE.length`

There are `POSSIBLE.length` possibilities for the chosen number.

The possibilities are `toSentence(POSSIBLE)`.

There `WANTED_COUNT == 1 ? "is only" : "are"` `plural(WANTED_COUNT, "favorable outcome")`: `toSentence(WANTED_LIST)`.

The probability is `\displaystyle ``fractionSimplification(WANTED_COUNT, POSSIBLE.length)`

.