Answer: 0.88
Step-by-step 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 :-
[tex]\text{P(A)}=\dfrac{16}{64}[/tex]
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 :-
[tex]P(A\cap B)}=\dfrac{14}{64}[/tex]
The conditional probability of B , given that A is given by :-
[tex]P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow\ P(B|A)=\dfrac{\frac{14}{64}}{\frac{16}{64}}\\\\\Rightarrow\ P(B|A)=\dfrac{14}{16}=0.875\approx0.88[/tex]
Hence, the probability that the child spends less than 1 hour per day on electronics =0.88
