Question

The relative frequency table shows the results of a survey in which parents were asked how much time their children spend playing outside and how much time they spend using electronics. Given that a child spends at least 1 hour per day outside, what is the probability, rounded to the nearest hundredth if necessary, that the child spends less than 1 hour per day on electronics?

Answer 1

The required probability is 88% or 0.88

Explanation:

Consider the provided table.

Let X represents the number of children spend at least 1 hour/day outside and number of children spend less than 1 hour/day on electronics respectively.

The total number of children that spend at least an hour a day outside is 16.

Probability of a child spends at least 1 hour per day outside is:
`P(X)=(16)/(64)`

Total number of children who spend less than 1 hour/day on electronics and spend at least 1 hour per day outside = 14

Probability of a child spends less than 1 hour per day on electronics and at least 1 hour per day outside is:
`P(X\cap Y)}=(14)/(64)`

According to conditional probability:

`P(Y|X)=(P(X\cap Y))/(P(X))\\P(Y|X)=((14)/(64))/((16)/(64))\\P(Y|X)=(14)/(16)\approx0.88`

Hence, the required probability is 88% or 0.88

Answer 2

We know there is a total of 16 children that spend at least an hour a day outside.

14 of those children spend at less than an hour a day on electronics.

So our fraction will be 14/16.

14/16 = 87.5%

Rounded = 88%

Hope this helped. Have a great day!