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

total 16 children spend at least 1 hour/day outside
and 14 out of 16 spend less than 1 hour/day using electronics

so 14/16 = 0.875 = 87.5% = 88%(rounded to nearest percent)

answer
88%

Answer 2

Answer: 0.88

Explanation:

Let A denotes the number of children spend at least 1 hour per day outside and B denote the number of children spend less than 1 hour per day on electronics .

From the given relative frequency table , we have

The total number of children spend at least 1 hour per day outside = 16

Probability of a child spends at least 1 hour per day outside is given by :-

`\text{P(A)}=(16)/(64)`

The total number of children spend less than 1 hour per 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 given by :-

`P(A\cap B)}=(14)/(64)`

The conditional probability of B , given that A is given by :-

`P(B|A)=(P(A\cap B))/(P(A))\\\\\Rightarrow\ P(B|A)=((14)/(64))/((16)/(64))\\\\\Rightarrow\ P(B|A)=(14)/(16)=0.875\approx0.88`

Hence, the probability that the child spends less than 1 hour per day on electronics =0.88