In Fig $$15.3$$ shown a disc on which a player spins an arrow twice. The function $$\dfrac{a}{b}$$ is formed, where a is the number of sector on which arrow stops on the first spin and $$b$$ is the number of the sector in which the arrow stops on second spin. On each spin, each sector has equal chance of selection by the arrow. Find the probability that the fraction $$\dfrac{a}{b}>1$$
For $$a/b > 1$$, when $$a=1, b$$ can not take any value,
$$a=2, b$$ can take $$1$$ value,
$$a=3, b$$ can take $$2$$ value,
$$a=4, b$$ can take $$3$$ values,
$$a=5, b$$ can take $$4$$ values,
$$a=6, b$$ can take $$5$$ values.
Total possible outcomes $$=36$$
$$\therefore P(a/b > 1)=\dfrac{1+2+3+4+5}{36}=\dfrac{15}{36}$$ or $$\dfrac{5}{12}$$
In fig. disc on which a player spins an arrow twice. The function is formed, where is the number of the sector on which the arrow stops on the first spin and is the number of the sector in which the arrow stops on the second spin. On each spin, each sector has an equal chance of selection by the arrow. Find the probability that the fraction .
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A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers (see Fig.) ,and these are equally likely outcomes. What is the probability that it will point at an odd number?
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $ 1,2,3,4,5,6,7,8$ (see Fig.) ,and these are equally likely outcomes. What is the probability that it will point at $ 8$?
4B