### Archive

Archive for the ‘Probability’ Category

## Questions: Probability And and OR

A bag contains 4 gold(G) balls and 7 silver(S) balls.  A ball is taken out its colour is noted and then it is put back.  Then this is repeated a second time.

What is the:

$P(G,G)$

$P(S,S)$

$P(\mbox{exactly one G})$

You have a biased die.  From a single roll the P(3)=0.1 and P(5)=0.4.  If the die is rolled twice what is the:

$P(3,3)$

$P(5,5)$

$P(\overline{3},\overline{3})$

$P(\mbox{exactly one 5})$

Becky wakes up one morning and is really tired and does not want to open her eyes.  In her draw she has 5 white(W) socks and 9 blue polka(P) dot socks, the two odd ones have gone to that place where all odd socks go.  She takes out a sock and puts it on and then takes out another to place on the other foot.  What is the:

$P(W then P)$

$P(P then W)$

$P(pair)$

Nicola and Louise are playing a very dangerous game and it is winner takes all.  In a bag there are 5 chocolates, 4 are praline flavour and 1 is a disgusting orange flavour.  To look at they are identical.  Whoever eats the orange chocolate loses.  Nicola is to take the first two chocolates and louise is to take the next two.  What is the:

P(Nicola looses)

P(Its a draw)

P(Louise looses)

Categories: Probability

## Probability: An event not happening

Consider the situation of a bag with 3 gold(G) balls and 2 silver(S) balls.

What is:

• P(G)?
• P(not G)?

$P(G)=\frac{3}{5}$, 3 gold balls and 5 balls in total.

$P(\text{not }G)=\frac{2}{5}$, 2 not gold (silver) balls and 5 balls altogether.

The important point to notice here is that the number of gold balls + number of not gold balls is equal to the total number of balls.

$\implies P(\text{not }A)=1-P(A)$

#### Notation

We can write P(not A) as $P(\overline{A})$, where the over bar means not.

Examples

If the $P(A)=0.7$, what is $P(\overline{A})$?

$P(\overline{A})=1-P(A)=1-0.7=0.3$

If the $P(win)=15\%$, what is $P(\overline{win})$

$P(\overline{win})=100\%-P(A)=100\%-15\%=85\%$

If the $P(blue)=\frac{4}{15}$, what is $P(\overline{blue})$

$P(\overline{blue})=1-P(blue)=1-\frac{4}{15}=\frac{11}{15}$

Categories: Probability

## Probability: AND & OR rule

### The ‘OR’ rule

The ‘OR’ rule states $P(A\text{ or } B)=P(A)+P(B)$

This makes sense since the number of successes(NoS) = (NOS of A) + (NoS of B)

The number of outcomes has not changed.

So $P(A\mbox{ or } B)=\frac{\mbox{(NOS of A) + (NoS of B)}}{\mbox{possible outcomes}}=\frac{\mbox{(NOS of A)}}{\mbox{possible outcomes}}+\frac{\mbox{(NOS of A)}}{\mbox{possible outcomes}}$

$\implies P(A\mbox{ or } B)=P(A)+P(B)$

Examples

A bag contains a red ball, 5 blue ball and 4 green balls.

1) P(red)

2) P(blue)

3) P(green)

4) P(red or green)

5) P(red or green or blue)

1) $P(red)=\frac{1}{10}$

2) $P(blue)=\frac{5}{10}$

3) $P(green)=\frac{4}{10}$

4) $P(\mbox{red or green})=P(red)+P(green)=\frac{1}{10}+\frac{4}{10}=\frac{5}{10}$

5) $P(\mbox{red or green or blue})=P(red)+P(green)+(blue)=\frac{1}{10}+\frac{4}{10}+\frac{5}{10}=1$

### The ‘AND’ rule

Here we are considering the probability of something happening and then something else happening.

Let’s consider a die rolled twice.

What is the P(two sixes)?

Well we will get a ‘6’ $\frac{1}{6}$ of the time.  Of these times we will get another ‘6’ $\frac{1}{6}$ of the time.

So $P(\mbox{two sixes})=\frac{1}{6}\text { of }\frac{1}{6}=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$

So $P(\mbox{A and B})=P(A)\times P(B)$

Examples

A bag contains 4 red(R) balls and 3 black(B) balls.  I take out a ball, look at it and then put it back.  I then take out another ball and record its colour.  What is the:

1. P(R,R) $\rightarrow$ means getting two reds.
2. P(R then B)
3. P(R,B) $\rightarrow$ means getting a red and blue in any order. i.e. red then blue or blue then red

$P(R)=\frac{4}{7}$

$P(B)=\frac{3}{7}$

So

$P(R,R)=\frac{4}{7} \times\frac{4}{7} =\frac{16}{49}$

$P(\mbox{R then B})=\frac{4}{7}\times \frac{3}{7}=\frac{12}{49}$

$P(\mbox{R, B})=P(\mbox{R then B}) + P(\mbox{B then R})$

$P(\mbox{B then R})=\frac{3}{7}\times \frac{4}{7}=\frac{12}{49}$

$P(R,B)=\frac{12}{49}+\frac{12}{49}=\frac{24}{49}$

#### Questions

Categories: Probability

## Questions: Equilikely outcomes

1.

Freya has a bag with 3 red balls and 2 green balls.  A ball is taken out and its colour is recorded.

What is P(red)?

The ball she took out was red.  Freya keeps the ball, she thinks it is lucky.

What is the probability that the next ball is red as well?

The ball she took out was green.  Freya didn’t like this ball, it reminded her of custard, so she flushed it down the toilet.

What is the probability the next ball is red?

2.

Inhoo was watching butterflies landing on a flower in a garden.  There were 7 white butterflies (Inhoo’s favourite) and 3 red butterflies.

What is the probability that the next butterfly to land on a flower is a white one?

In fact a red butterfly landed on the flower.  Inhoo was so angry set caught the butterfly and threw it in a nearby spiders web.

What is the probability that the next butterfly to land on a flower is a white one?

Again a red butterfly landed on the flower.  Inhoo almost cried, but she gathered herself together, caught the butterfly and put it in a jar.

What is the probability that the next butterfly to land on a flower is a white one?

Yet again a red butterfly landed on the flower.  This time Inhoo took no chances, she caught the butterfly and ate it there and then.

What is the probability that the next butterfly to land on a flower is a white one?

3.

From a pack of cards what is:

P(black)?

P(king)?

P(2 or 3)?

P(red ace)?

4.

Lydia has a set of cards.  On them she has either drawn a heart or a dagger.  She has drawn 6 daggers.  If $P(\text{dagger})=\frac{2}{5})$, how many hearts did she draw?

Categories: Probability

## Probability: Equilikely outcomes

The probability of an event happening is how likely it is to happen or how frequently it happens.

In a perfectly fair world if a die was thrown a five would come up every six throws.  That means that the probability of getting a 5 = $\frac{1}{6}$.

Generally the probability of something happening $=\frac{\text{possible successes}}{\text{possible outcomes}}$

#### Notation

The probability of A happening can be written as P(A).

So P(red) could be what is the probability of getting red?

Examples

What is the probability of rolling a 3 with a die?

There is 1 possible success (getting 3)

There are 6 possible outcomes (1, 2, 3, 4, 5 or 6)

So $P(3)=\frac{1}{6}$

When rolling a die what is the probability of getting a square number?There are 2 possible successes (1 or 4)There are 6 possible outcomes (1, 2, 3, 4, 5 or 6)

So $P(\text{square})=\frac{2}{6}=\frac{1}{3}$

When playing cards what is the probability of getting a red heart?There are 13 possible successes (13 hearts)There are 52 possible outcomes (52 cards in total)

So $P(\text{heart})=\frac{13}{52}=\frac{1}{2}$

A bag contains 5 red discs and 7 blue discs.I take out a red disc and keep it.  What is the probability the next disc is also red?There are 4 successes (there are only 4 red discs left as we have taken one our already)

There are 11 possible outcomes (there are only 11 discs left now)

So $P(\text{red})=\frac{4}{11}$

Important note: Probabilities can only be given as fractions, decimals or percentages

Questions

Categories: Probability