Question

In a single cast with two dice, the odds against drawing 7 is
a. 1/6
b. 1/12
c. 5:1
d. 1:5

Answer 1

In this type of question we have to use the concept of probability. We know that probability is the measure of possibility of occurrence of an event. Odds are used to describe the chance of an event occurring. The odds are the ratios that compare the number of ways the event can occur with the number of ways the event cannot occur. The odds against the event are the ratio of the number of ways that an outcome cannot occur compared to in how many ways it can occur. Hence, we can write, $\text{Odds against the event = }\dfrac{\text{No}\text{. of ways in the event cannot occur}}{\text{No}\text{. of ways in the event occur}}$

Complete step by step solution:
Now here we have to find the odds against drawing 7 in a single cast with two dice.
We know that if two dice are thrown simultaneously then possible outcomes are

& \left( 1,6 \right),\left( 1,5 \right),\left( 1,4 \right),\left( 1,3 \right),\left( 1,2 \right),\left( 1,1 \right) \\\ & \left( 2,6 \right),\left( 2,5 \right),\left( 2,4 \right),\left( 2,3 \right),\left( 2,2 \right),\left( 2,1 \right) \\\ & \left( 3,6 \right),\left( 3,5 \right),\left( 3,4 \right),\left( 3,3 \right),\left( 3,2 \right),\left( 3,1 \right) \\\ & \left( 4,6 \right),\left( 4,5 \right),\left( 4,4 \right),\left( 4,3 \right),\left( 4,2 \right),\left( 4,1 \right) \\\ & \left( 5,6 \right),\left( 5,5 \right),\left( 5,4 \right),\left( 5,3 \right),\left( 5,2 \right),\left( 5,1 \right) \\\ & \left( 6,6 \right),\left( 6,5 \right),\left( 6,4 \right),\left( 6,3 \right),\left( 6,2 \right),\left( 6,1 \right) \\\ \end{aligned} \right\\}$$ $$\Rightarrow n\left( \text{S} \right)\text{ = }36$$ Now, let us consider, E be the event of getting 7 $$\begin{aligned} & \Rightarrow \text{E = }\left\\{ \left( 1,6 \right),\left( 2,5 \right),\left( 3,4 \right),\left( 4,3 \right),\left( 5,2 \right),\left( 6,1 \right) \right\\} \\\ & \Rightarrow n\left( \text{E} \right)\text{ = 6} \\\ \end{aligned}$$ And hence we can say that, $$\overline{\text{E}}$$ represent the event of not getting 7 $$\begin{aligned} & \Rightarrow n\left( \overline{\text{E}} \right)=n\left( \text{S} \right)-n\left( \text{E} \right) \\\ & \Rightarrow n\left( \overline{\text{E}} \right)=36-6 \\\ & \Rightarrow n\left( \overline{\text{E}} \right)=30 \\\ \end{aligned}$$ Hence, $$\text{Odds against the event E = }\dfrac{\text{No}\text{. of ways in the event E cannot occur}}{\text{No}\text{. of ways in the event E occur}}$$ $$\text{Odds against the event E = }\dfrac{\text{30}}{\text{6}}$$ $$\text{Odds against the event E = }\dfrac{5}{1}$$ **Thus the odds against drawing 7 in a single cast with two dice is $$\dfrac{5}{1}$$ that is $$5:1$$ And hence option (c) is correct. ** **Note:** In this type of question students may make mistakes in calculation of odds against the event. Students have to take care as we have to calculate odds against the event and not in favour of the event as both are totally different.