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

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!

Answer 2

Answer:

The required probability is 88% or 0.88

Step-by-step 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: $P(X)=\dfrac{16}{64}$

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: $P(X\cap Y)}=\dfrac{14}{64}$

According to conditional probability:

$P(Y|X)=\dfrac{P(X\cap Y)}{P(X)}\\P(Y|X)=\dfrac{\frac{14}{64}}{\frac{16}{64}}\\P(Y|X)=\dfrac{14}{16}\approx0.88$

Hence, the required probability is 88% or 0.88