Question

In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is $ \dfrac{5}{6} $ . What is the probability that he will knock down fewer than 2 hurdles?

Answer 1

Hint:
We start solving the problem by assigning the variables to the probability that the player clears the hurdle and the probability that the player knockdown. We then find the probability that the player can knockdown due to a hurdle. We then find the required probability by finding the number of ways to choose the hurdles which were to be cleared and then multiply it with the respective probabilities to get the required answer.

Complete step by step answer:
According to the problem, we are given that a player has to cross 10 hurdles in a hurdle race. We need to find the probability that he will knock down fewer than 2 hurdles if the probability that he will clear each hurdle is $ \dfrac{5}{6} $.
Let us assume the probability that the player clears the hurdle be ‘p’ and the probability that the player knockdown be ‘q’.
So, we have $ p=\dfrac{5}{6} $ .
Now, let us find the probability that he will knockdown.
So, we have $ p+q=1\Leftrightarrow q=1-\dfrac{5}{6}=\dfrac{1}{6} $ .
Now, we know that the player has to clear at least hurdles in order to knock fewer than 2 hurdles.
So, the required probability will be $ P\left( X=9 \right)+P\left( X=10 \right) $ .
Here we need to find the number of ways to choose the hurdles that were cleared by the player and then multiply it with the respective probability.
 $ \Rightarrow P\left( X=9 \right)+P\left( X=10 \right)=\left( {}^{10}{{C}_{9}}{{p}^{9}}q \right)+\left( {}^{10}{{C}_{10}}{{p}^{10}} \right) $ .
 $ \Rightarrow P\left( X=9 \right)+P\left( X=10 \right)=\left( {}^{10}{{C}_{9}}{{\left( \dfrac{5}{6} \right)}^{9}}\left( \dfrac{1}{6} \right) \right)+\left( {}^{10}{{C}_{10}}{{\left( \dfrac{5}{6} \right)}^{10}} \right) $ .
 $ \Rightarrow P\left( X=9 \right)+P\left( X=10 \right)=\left( \dfrac{10\times {{5}^{9}}}{{{6}^{10}}} \right)+\left( \dfrac{1\times {{5}^{10}}}{{{6}^{10}}} \right) $ .
 $ \Rightarrow P\left( X=9 \right)+P\left( X=10 \right)=\dfrac{{{5}^{9}}}{{{6}^{9}}}\left( \dfrac{10}{6}+\dfrac{5}{6} \right) $ .
 $ \Rightarrow P\left( X=9 \right)+P\left( X=10 \right)=\dfrac{{{5}^{9}}}{{{6}^{9}}}\left( \dfrac{15}{6} \right) $ .
 $ \Rightarrow P\left( X=9 \right)+P\left( X=10 \right)=\dfrac{{{5}^{10}}}{2\times {{6}^{9}}} $ .
 $ \therefore $ We have found the required probability as $ \dfrac{{{5}^{10}}}{2\times {{6}^{9}}} $ .

Note:
Here we have assumed that the player must cross all the 6 hurdles once he starts running and does not quit in the middle of the game. We assumed that clearing or knocking each hurdle is the independent event for solving this problem. We should not make calculation mistakes while solving this problem. Similarly, we can expect a problem to find the probability that the player clears exactly 5 hurdles.