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