91Ó°ÊÓ

We are given the scores and the probability that one probability of score is missing, and we have to find out the probability that he or she has earned a score of at least three.

In the given table, scores and probabilities are given but one probability is missing, as the sum of total probabilities is one.

So, we get $P\left(T\right)=1$where P(T) is the sum of all probability given

Putting the values we get,$0.235+0.155+0.249+0.217+x=1$

which states x$=0.144$,

So, now we have all probabilities.

The probability of scoring at least three will be the sum of all probabilities greater than$3$including itself.

P($3$)$=0.249$, P($4$)$=0.217$,and P($5$)$=0.144$.

As, the P(T)$=P\left(3\right)+P\left(4\right)+P\left(5\right)$

On solving we get ,$P\left(T\right)=0.610$.

Hence, the probability of scoring at least three is$0.610$.