Question

If the odds against Deborah winning first prize in the chess tournament are 2 to 11, what is the probability of the event that she will win first prize? the probability of Deborah winning first prize is ______. (type an integer or a fraction).

Answer 1

The question says that the odds against Deborah winning first prize in the chess tournament are 2 to 11. These are the odds of her not winning.

Odds of an event is related to the probability of the same event occuring by the formula:

`\begin{gathered} O=(p)/(1-p) \\ \\ \text{where,} \\ O=\text{Odds} \\ P=\text{probability} \\ \\ \end{gathered}`

Therefore, to calculate the probability of the event occurring, we simply make p the subject of the formula:

`\begin{gathered} p=(O)/(O+1) \\ \\ \text{where,} \\ p=\text{probability} \\ O=\text{Odds} \end{gathered}`

Now that we have the formula for probability (p) in terms of odds (O), we can proceed to find the probability of Deborah NOT winning first prize. This is done below:

`\begin{gathered} O=(2)/(11)\text{ (or 2 to 11)} \\ \\ p=(O)/(O+1) \\ \\ O+1=(2)/(11)+1=(13)/(11)_{} \\ \\ \therefore p=(2)/(11)/(13)/(11) \\ p=(2)/(11)*(11)/(13)\text{ (11 crosses out)} \\ \\ \therefore\text{Probability of Deborah NOT winning is:}(2)/(13) \end{gathered}`

The question asks us to find the probability that Deborah wins the first prize. This is the direct opposite event or the mutually exclusive event to her NOT winning.

The relationship between event (P) and its opposite, event (Q) is:

`P=1-Q`

Therefore, to find the probability (Q) of Deborah winning first prize is:

`\begin{gathered} (2)/(13)=1-Q \\ \\ \text{make Q the subject of the formula by adding Q to both sides and} \\ \text{subtracting 2/13 from both sides} \\ \\ (2)/(13)-(2)/(13)+Q=1-Q+Q-(2)/(13) \\ Q=1-(2)/(13) \\ \\ \therefore Q=(11)/(13) \end{gathered}`

Therefore the probability of Deborah winning the first prize is: 11/13