Question

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

Answer 1

Crossing hurdles follows Bernoulli trial method.
Here, first we have to identify whether the given probability is of positive outcome or negative outcome and find the outcome that is not given. Then we put them in the binomial distribution formula and find out the necessary probability that is needed to be obtained. The positive and negative outcomes are named P, Q respectively according to the given problem.

Complete step by step solution:
First
Let $X:$ be the number of hurdles knocked down by the player.
Since crossing hurdles follows Bernoulli trial.
So, X has the binomial distribution as
$P(X = x) = {}^n{C_x}{q^{n - x}}{p^x}$
From the question, we are given that
$n =$ number of hurdles = 10. Also
The probability that the player will clear hurdle $= \dfrac{5}{6}$
So, $q = \dfrac{5}{6}$
Hence,
$p = 1 - q = 1 - \dfrac{5}{6} = \dfrac{1}{6}$
Now, the probability of knocking down “X” number of hurdles

$P(X = x) = {}^{10}{C_x}{\left( {\dfrac{5}{6}} \right)^{10 - x}}{\left( {\dfrac{1}{6}} \right)^x}$.
Since, we have to find the probability of knocking down hurdles fewer than 2.
Since knocking down no hurdles and also knocking down 1 hurdle is to be considered and added to get the required probability.
$P(X < 2) = P(X = 0) + P(X = 1)$
Which means
P(Knocking down fewer than 2 hurdles ) = P( knocking down zero hurdles )+ P(knocking down one hurdle).

$$= {}^{10}{C_0}{\left( {\dfrac{5}{6}} \right)^{10 - 0}}{\left( {\dfrac{1}{6}} \right)^0} + {}^{10}{C_1}{\left( {\dfrac{5}{6}} \right)^{10 - 1}}{\left( {\dfrac{1}{6}} \right)^1} \\\ = 1 \times 1 \times {\left( {\dfrac{5}{6}} \right)^{10}} + 10 \times \left( {\dfrac{1}{6}} \right) \times {\left( {\dfrac{5}{6}} \right)^9} \\\ = {\left( {\dfrac{5}{6}} \right)^{10}} + \dfrac{{10}}{6}{\left( {\dfrac{5}{6}} \right)^9} \\\ = \dfrac{5}{6}{\left( {\dfrac{5}{6}} \right)^9} + \dfrac{{10}}{6}{\left( {\dfrac{5}{6}} \right)^9} \\\ = \left( {\dfrac{5}{6} + \dfrac{{10}}{6}} \right){\left( {\dfrac{5}{6}} \right)^9} \\\ = \left( {\dfrac{{15}}{6}} \right){\left( {\dfrac{5}{6}} \right)^9} \\\ = \dfrac{5}{2}{\left( {\dfrac{5}{6}} \right)^9} \\\ = \dfrac{{{5^{10}}}}{{2 \times {6^9}}} \\\$$

Hence, we have found out the required probability of the player knocking down less than 2 hurdles

Note: Recognition of whether given probability is the positive outcome or the negative outcome based on the asked probability is very important and should be done carefully. Also, the final is not needed to be completely calculated, unless you don’t need a calculator. Also, we have to recognize what type of trial the question follows.