Hi! I'm Jack and I'm a first year student studying a masters in mathematics at the University of Exeter. My passion for maths has always been a big part of my life as I took part in competitions and extra-curricular activities to do with mathematics throughtout my school career. I hope to bring this passion for mathematics, and by association, my love for chemistry, to my tuttees and help them to not only get the answers right but also understand **why** they are right.

The sessions will be as tailormade as you require, focussing on topics that you have told me you struggle on and will include questions of multiple degrees of difficulty.

If you have any questions, please do not hesitate to contact me or book a **free** Meet the Tutor session!

I look forward to hearing from you :)

Hi! I'm Jack and I'm a first year student studying a masters in mathematics at the University of Exeter. My passion for maths has always been a big part of my life as I took part in competitions and extra-curricular activities to do with mathematics throughtout my school career. I hope to bring this passion for mathematics, and by association, my love for chemistry, to my tuttees and help them to not only get the answers right but also understand **why** they are right.

The sessions will be as tailormade as you require, focussing on topics that you have told me you struggle on and will include questions of multiple degrees of difficulty.

If you have any questions, please do not hesitate to contact me or book a **free** Meet the Tutor session!

I look forward to hearing from you :)

No DBS Check

When Dave throws the ball, it can either go into the bucket, or miss. The probability of the ball going into the bucket is 0.4 and each throw is independent of eachother so this probability is the same for each throw. As the only other event that can occur is Dave missing, the probability of him missing must be

P(miss) = 1 - 0.4 = 0.6

For the described event to occur, Dave must get the ball in the bucket twice, and miss twice. To calculate the probability of multiple independent events occuring, we multiply the probabilities, commonly referred to as

P(A n B) = p(A) x p(B)

Thus,

0.4 x 0.4 x 0.6 x 0.6

This can be written in a simpler form as

0.4^{2 }x 0.6^{2}

*This solution gives us the probability that Dave will miss twice and then success twice, but the question does not specify an order.*

The only factor left to consider is **combinations**. This is the mathematical way of considering how many ways an event can occur. In this event, we need **2** of the **4** throws to be misses (or successes), so we must calculate how many different ways this could occur. We can do this in different ways:

__Listing__

success success miss miss

success miss success miss

success miss miss success

miss success success miss

miss success miss success

miss miss success success

*This method gives us the (correct) answer of 6, but it is easy to make a mistake and miss one of the combinations out.*

__Pascal's Triangle__

You may be familiar with Pascal's triangle from the binomial theorem or many popular maths puzzles, it is a triangle of numbers such that every number is the sum of the two numbers above it, starting with one.

0 1

1 1 1

2 1 2 1

3 1 3 3 1

4 1 4 6 4 1

5 1 5 10 10 5 1

(notice the row numbers on the left begin with row 0)

Using the triangle, we note that we are looking for the number of ways to arrange **2** events within **4**. So we look at place** 2** on row **4**. Musch like with rows, the places within the rows, start with the 0th, so the 2nd term in row 4 is **6.**

*This method has also given us the correct answer of 6. This method is, again complicated, and ceases to be helpful with larger numbers, but does help in understanding why the final and most efficient method works.*

__Choose__

Choose is a mathematical method used in combinations. Found on all scientific calculators it will often be displayed in the format

nCr

This is, in effect the more mathematical version of the Pascal's Triangle method. Again we are looking for the number of ways we can arrange 2 events within 4. So, in your calculator, you type

4C2

and the answer will be 6.

*This is, in my opinion the best method as the only thing that can be confusing is the order of the numbers. This is only a small problem as the calculator will only work with the correct order as the other is impossible. And you can always remember it as 'out of 4 events we are choosing 2' so its 4 choose 2!*

Now we simply multiply our probability from earlier (which gives the probability of one combination occuring) by the number of combinations.

0.4^{2 }x 0.6^{2} x 6 = 0.3456

When Dave throws the ball, it can either go into the bucket, or miss. The probability of the ball going into the bucket is 0.4 and each throw is independent of eachother so this probability is the same for each throw. As the only other event that can occur is Dave missing, the probability of him missing must be

P(miss) = 1 - 0.4 = 0.6

For the described event to occur, Dave must get the ball in the bucket twice, and miss twice. To calculate the probability of multiple independent events occuring, we multiply the probabilities, commonly referred to as

P(A n B) = p(A) x p(B)

Thus,

0.4 x 0.4 x 0.6 x 0.6

This can be written in a simpler form as

0.4^{2 }x 0.6^{2}

*This solution gives us the probability that Dave will miss twice and then success twice, but the question does not specify an order.*

The only factor left to consider is **combinations**. This is the mathematical way of considering how many ways an event can occur. In this event, we need **2** of the **4** throws to be misses (or successes), so we must calculate how many different ways this could occur. We can do this in different ways:

__Listing__

success success miss miss

success miss success miss

success miss miss success

miss success success miss

miss success miss success

miss miss success success

*This method gives us the (correct) answer of 6, but it is easy to make a mistake and miss one of the combinations out.*

__Pascal's Triangle__

You may be familiar with Pascal's triangle from the binomial theorem or many popular maths puzzles, it is a triangle of numbers such that every number is the sum of the two numbers above it, starting with one.

0 1

1 1 1

2 1 2 1

3 1 3 3 1

4 1 4 6 4 1

5 1 5 10 10 5 1

(notice the row numbers on the left begin with row 0)

Using the triangle, we note that we are looking for the number of ways to arrange **2** events within **4**. So we look at place** 2** on row **4**. Musch like with rows, the places within the rows, start with the 0th, so the 2nd term in row 4 is **6.**

*This method has also given us the correct answer of 6. This method is, again complicated, and ceases to be helpful with larger numbers, but does help in understanding why the final and most efficient method works.*

__Choose__

Choose is a mathematical method used in combinations. Found on all scientific calculators it will often be displayed in the format

nCr

This is, in effect the more mathematical version of the Pascal's Triangle method. Again we are looking for the number of ways we can arrange 2 events within 4. So, in your calculator, you type

4C2

and the answer will be 6.

*This is, in my opinion the best method as the only thing that can be confusing is the order of the numbers. This is only a small problem as the calculator will only work with the correct order as the other is impossible. And you can always remember it as 'out of 4 events we are choosing 2' so its 4 choose 2!*

Now we simply multiply our probability from earlier (which gives the probability of one combination occuring) by the number of combinations.

0.4^{2 }x 0.6^{2} x 6 = 0.3456