Answer to Question #166682 in Statistics and Probability for Stephanie

Question

Only 27% of U.S adults get enough leisure time to exercise to achieve cardiovascular fitness. Choose 3 adults at random. Find the probability that A: All 3 get enough daily exercise B: At least 1 of the 3 gets enough exercise.

Answer 1

Let us consider U.S adults get enough time to exercise as a success.

Then probability of success is $\frac{27}{100}$ and probability of failure is $\frac{73}{100}$.

$\therefore p=\frac{27}{100}$ and $q=\frac{73}{100}$.

Now we know that for binomial distribution probability of getting $r$ success from $n$ trial is $={^n}C_r.p^r.q^{(n-r)}$

(A) Here $n=3,r=3$

Therefore probability that all $3$ get daily enough exercise is $={^3}C_3.(\frac{27}{100})^3.(\frac{73}{100})^{(3-3)}=(\frac{27}{100})^3$ .

(B) First we find out of $3$ students no one get enough time to exercise.

Therefore here $n=3,r=0$

So probability that no one get enough exercise is $={^3}C_0.(\frac{27}{100})^0.(\frac{73}{100})^{(3-0)}=(\frac{73}{100})^3$ .

Therefore the required probability of getting atleast $1$ of the $3$ gets enough exercise is $=[1-(\frac{73}{100})^3]$ $=0.61$