Assessing risk in engineering, work and life
Assessing risk in engineering, work and life

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Assessing risk in engineering, work and life

3.3 Probability in everyday life

Here are some interesting probability examples. You may like to try working out the answers yourself to enhance your understanding, or you can use them to entertain your family and friends!

Activity 9

Timing: Allow approximately 25 minutes.

Winning the UK National Lottery

The UK National Lottery operates a number of ‘games’ each week. In the main game, participants choose six different numbers from 1 to 59. When the draw is made, six numbered balls are selected at random from a pool of 59 balls (ignoring the seventh ‘Bonus Ball’). If a participant’s selection of numbers matches the six numbers drawn (although the order of the numbers may be different), they win the jackpot, which can be several million pounds. Smaller prizes can be won by matching two or more numbers.

What is the probability of any single entry winning the jackpot?

Answer

The first time a ball is selected, there are 59 to choose from and 6 of these will match the numbers chosen by the participant, so the probability of getting a match is six divided by 59

When the second ball is drawn, there are 58 balls remaining and five of those will match one of the participant’s remaining numbers. This means that the probability of getting a match with the second ball is five divided by 58

So overall, the probability of matching two numbers is

equation left hand side six divided by 59 times prefix multiplication of five divided by 58 equals right hand side 30 divided by 3422 postfix times

which is approximately one divided by 114. If the calculation is continued in this way, the chance of matching all six balls is found to be given by

equation left hand side six divided by 59 times prefix multiplication of five divided by 58 equals right hand side 30 divided by 3422 postfix times
equation left hand side six divided by 59 times prefix multiplication of five divided by 58 times prefix multiplication of four divided by 57 times prefix multiplication of three divided by 56 multiplication two divided by 55 multiplication one divided by 54 equals right hand side 720 divided by 32 postfix times 441 postfix times 381 postfix times 280 postfix times equation left hand side equals right hand side one divided by 45 postfix times 057 postfix times 474 postfix times

So there is a probability of about 1 in 45 million that any one National Lottery ticket will win the jackpot. Of course, millions of tickets are sold each week, so it is not suprising that, in most weeks, one or more people win the jackpot.

Monkeys writing Shakespeare

There is an old saying that an infinite number of monkeys sitting in front of (an infinite number of) typewriters would eventually type all the great works of literature, including all of Shakespeare’s plays.

If we update this saying and take an infinite number of monkeys that enjoy pressing keys and give them access to monkey-friendly keyboards that contains just 26 keys (one for each letter of the alphabet) what is the probability of one monkey typing the word ‘MACBETH’ in seven keystrokes?

Answer

Unlike drawing the lottery numbers, the order of the monkey’s selection does matter and the same key can be pressed more than once. Taking the first letter, there is a 1 in 26 chance that the monkey hits the ‘M’ key. Similarly, there is a 1 in 26 chance that the second key is an ‘A’ so the chance of typing ‘MA’ in two keystrokes is given by

equation left hand side one divided by 26 times prefix multiplication of one divided by 26 postfix times equals right hand side one divided by 676

Continuing this line of reasoning, the chance of writing ‘MACBETH’ in seven keystrokes is given by

equation left hand side one divided by two times six super seven equals right hand side one divided by eight postfix times 031 postfix times 810 postfix times 176

So there is a probability of approximately 1 in 8 million that a monkey would write ‘MACBETH’ by hitting a typewriter keyboard seven times. Of course, this does not mean that it would not happen, just that it would be highly improbable. If the monkey were to type the meaningless sequence ‘KWCPETE’, the probability of that coming up would also have been 1 in 8 billion.

Two people sharing a birthday

If you are with a group of your friends, how big would the group have to be for there to be a reasonable (say 50%) probability that at least two of them were born on the same day of the year, but not necessarily in the same year?

When asked this, most people would vastly overestimate the size of the group, but if you think about this carefully you will come up with a result that may surprise you.

Answer

The first person in your group has a birthday on any one of 365 days (if we ignore leap years). The probability that the second person shares that birthday is one divided by 365 or, in other words, the probability that the second person’s birthday is on a different day is 364 divided by 365.

Now bring in a third person. The probability that they share a birthday with either of the first two people is two divided by 365, so the probability that they have a different birthday from either of the first two people is 363 divided by 365. Therefore the probability that the three people all have different birthdays is given by

equation sequence 364 divided by 365 times prefix multiplication of 363 divided by 365 postfix times equals 132 postfix times 860 divided by 133 postfix times 225 postfix times equals 0.997 postfix times prefix times of open t times o postfix times three postfix times s full stop f full stop close

Therefore, there is only a 1 − 0.997 = 0.003 (or 3 in 1000) chance that this is not the case and that at least two of the three people share the same birthday. This is perhaps not surprising.

Continuing down this path, it can be shown that by the time there are 23 people, the probability that they all have different birthdays is approximately 0.49, because

equation left hand side 364 times prefix multiplication of 363 times prefix multiplication of full stop full stop full stop multiplication 342 divided by 365 super 22 equals right hand side 0.493 postfix times open t times o postfix times three postfix times s full stop f full stop close postfix times

So the probability that this is not the case and at least two people share a birthday is 1 − 0.493 = 0.507. This means that there is a slightly better than 50% chance that at least two people share the same birthday.

If the size of the group continues to increase, by the time reaches 32 people there is a 75% chance that at least two people will share a birthday, because

equation left hand side 364 times prefix multiplication of 363 times prefix multiplication of full stop full stop full stop multiplication 333 divided by 365 super 31 equals right hand side 0.247 postfix times open t times o postfix times three postfix times s full stop f full stop close postfix times

so the probability that this is not the case and at least two people share a birthday is 1 − 0.247 = 0.753 or 75%.

T193_1

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