Lesson video

- Hi, I'm Mrs. Dennett, and in this lesson, we're going to be calculating probabilities from equally likely events.

The next question, Mary-Kate is selecting a doughnut at random from a box.

We have to work out some probabilities.

What is the probability that she selects a doughnut with a hole in the middle? There are six doughnuts with a hole in the middle, out of a total of 12 doughnuts in the box.

We write this as a fraction 6/12, but we can simplify this to 1/2.

Divide in the numerator and denominator by six.

We could also write this as a percentage as 1/2 is 50 hundreds, so 50%.

We could also write probability as decimals.

So 0.

5 would also be an acceptable answer for this question.

However, we cannot just write six as this is not a probability.

Six just represents the number of doughnuts with a hole in them in the box.

So we cannot just write six, neither can we write 6:12 as this is a ratio and we never write probabilities as ratios.

Next.

What is the probability that Mary-Kate picks a doughnut with blue icing? There are two doughnuts with blue icing, so we write 2/12 and this can be simplified to 1/6.

What is the probability that Mary-Kate picks a doughnut with pink icing? There are four doughnuts out of a total of 12, which have pink icing.

So the probability is 4/12.

We can write this as a fraction and simplify it, diving the numerator and denominator by four to get 1/3.

How else could we have written this probability? Well, we could have written this as a decimal, 0.

3 recurring or a percentage, 33.

3 recurring percent.

We cannot just write 4 though as this is just the number of doughnuts and not a probability.

We shouldn't be using ratios either.

Ratios are not used for probabilities.

Finally, we want the probability that the doughnut does not have brown icing.

We can see that two doughnuts do have brown icing and 10 don't.

So I can write 10/12 and simplify this to 5/6.

Here's a question for you to try.

Pause the video to complete the task and restart when you were finished.

Here are the answers.

There are six possible equally likely outcomes from rolling a die.

These are one, two, three, four, five or six.

There is only one four, so the probability of rolling a four is 1/6.

Rolling a seven is impossible.

So the probability of rolling a seven is zero.

There are three odd numbers on a die, one, three, and five.

So it probability of rolling an odd number is 3/6 or 1/2.

There are five other possibilities other than getting a four when rolling a die.

We can get one, two, three, five, or six.

So the next answer is 5/6.

The multiples of three are three and six on the die.

This gives us a probability of 2/6 or 1/3.

Square numbers on the die are one and four.

So again, this gives us a probability of 2/6 or 1/3.

Finally, there are five numbers greater than one on the die.

So the probability of rolling a number more than one is 5/6 Earlier, we worked out that the probability of getting a doughnut with a hole in the middle is 1/2.

This is quite a long way of writing all of that out.

We can abbreviate this.

We often just represent probability with a P and report in the bracket the event we are focusing on.

So here the event is at the doughnut with a hole in it.

So I can write P for probability, put the event doughnut with a hole in brackets is equal to 1/2.

This is just a shorthand way of writing out a probability.

Here's a question for you to try.

Pause the video to complete the task and restart when you were finished.

Here are the answers.

For question two, there are two yellow counters and a total of six counters in the bag.

So we have a probability 2/6 which can simplify to 1/3.

For question three, there are 10 sections on the spinner each section is worth 1/10.

So we call a three sections red to start with.

Then we have to find an equivalent fraction to 1/5 with a denominator of 10.

Multiply numerator and denominator by two gives us 2/10.

So two sections need to be shared in blue.

Finally, I know that 0.

5 is 5/10 or 1/2.

So five sections of the spinner or 1/2 of the spinner must be green.

Here's another question for you to try.

Pause the video to complete this task and restart when you were finished.

Here are the answers.

We know that there are 12 counter in the box.

So straightaway we can see that there are 5/12 orange counters because the denominator of this fraction is 12.

For the yellow and white counters, we need to find equivalent fractions for 1/4 and 1/3, with denominators of 12 or we can find 1/4 of 12 and 1/3 of 12.

I'm going to use this method as I think it will be quicker.

1/4 of 12 is 12 divided by four, which is three.

So there are three yellow counters.

1/3 of 12 is 12 divided by three equal to four.

So there are full white counters.

For part B, we need to work out the probability of getting an orange counter and compare it to the probability of not getting an orange counter.

We can use our answer to part A because we know that there are five orange counters and there are seven counters that are not orange.

So it is less likely that we will get an orange counter.

Finally, for part C, we want to know how many counters we would have to select from the box before getting an orange counter.

There is a chance that we could pick all the yellow and white counters before getting an orange one.

There are seven of these, so it could be the eighth pick before we select an orange counter.

That's all for this lesson.

Thank you for watching.