# Struct rand::distributions::Bernoulli

source · `pub struct Bernoulli { /* private fields */ }`

## Expand description

The Bernoulli distribution.

This is a special case of the Binomial distribution where `n = 1`

.

## Example

```
use rand::distributions::{Bernoulli, Distribution};
let d = Bernoulli::new(0.3).unwrap();
let v = d.sample(&mut rand::thread_rng());
println!("{} is from a Bernoulli distribution", v);
```

## Precision

This `Bernoulli`

distribution uses 64 bits from the RNG (a `u64`

),
so only probabilities that are multiples of 2^{-64} can be
represented.

## Implementations§

source§### impl Bernoulli

### impl Bernoulli

source#### pub fn new(p: f64) -> Result<Bernoulli, BernoulliError>

#### pub fn new(p: f64) -> Result<Bernoulli, BernoulliError>

Construct a new `Bernoulli`

with the given probability of success `p`

.

##### Precision

For `p = 1.0`

, the resulting distribution will always generate true.
For `p = 0.0`

, the resulting distribution will always generate false.

This method is accurate for any input `p`

in the range `[0, 1]`

which is
a multiple of 2^{-64}. (Note that not all multiples of
2^{-64} in `[0, 1]`

can be represented as a `f64`

.)

source#### pub fn from_ratio(
numerator: u32,
denominator: u32
) -> Result<Bernoulli, BernoulliError>

#### pub fn from_ratio( numerator: u32, denominator: u32 ) -> Result<Bernoulli, BernoulliError>

Construct a new `Bernoulli`

with the probability of success of
`numerator`

-in-`denominator`

. I.e. `new_ratio(2, 3)`

will return
a `Bernoulli`

with a 2-in-3 chance, or about 67%, of returning `true`

.

return `true`

. If `numerator == 0`

it will always return `false`

.
For `numerator > denominator`

and `denominator == 0`

, this returns an
error. Otherwise, for `numerator == denominator`

, samples are always
true; for `numerator == 0`

samples are always false.