#### Answer

There is sufficient evidence to support that less than 1/3 of the challenges are succesful.

#### Work Step by Step

$H_{0}:p=1/3$. $H_{a}:p<1/3.$ $\hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=\frac{x}{n}=\frac{172}{611}=0.2815.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.2815-1/3}{\sqrt{1/3(1-1/3)/611}}=-2.72.$ The P is the probability of the z-score being less than -2.72, hence:P=1-0.0033. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.0033 is less than $\alpha=0.01$, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to support that less than 1/3 of the challenges are succesful.