Question

assume that when adults with smartphones are randomly selected 47% use them in meetings or classes if 6 adult smartphones users are randomly selected find the probability that at least four of them use their smartphones in meetings or classes

Answer 1

Given:

p = 47% = 0.47

n = 6

Required:

The probability that at least 4 are using their smartphones in meetings or classes.

Solution:

The binomial probability is defind as :

We will use this definition to solve for P at X = 4, X = 5 and X = 6 (since the probability of at least 4 users is asked)

`\begin{gathered} P(X=4)=(6!)/(4!(6-4)!)\cdot0.47^4\cdot(1-0.47)^(6-4)=(6!)/(4!2!)\cdot0.47^4\cdot0.53^2 \\ P=15\cdot0.47^4\cdot0.53^2=0.2056 \end{gathered}`
`\begin{gathered} P(X=5)=(6!)/(5!(6-5)!)\cdot0.47^5\cdot(1-0.47)^(6-5)=(6!)/(5!1!)\cdot0.47^5\cdot0.53^1 \\ \\ P=0.05\cdot0.47^5\cdot0.53^{}=6.0776\cdot10^4 \end{gathered}`
`\begin{gathered} P(X=6)=(6!)/(6!(6-6)!)\cdot0.47^6\cdot(1-0.47)^(6-6)=(6!)/(6!0!)\cdot0.47^6\cdot0.53^0 \\ P=1\cdot0.47^6\cdot1^{}=0.0108 \end{gathered}`