Whooping cough is a highly contagious bacterial infection...About 80% of unvaccinated children who are exposed to whooping cough will develop the infection, as opposed to only about 5% of vaccinated children.
1) Find the probability that exactly 2 out of 20 exposed children develop whooping cough if 17 children have been vaccinated and 3 has not been.
My solution for this question is: 17 children = 0.05 ^ 17 and 3 children = 0.8 ^ 3
2 out of 20 = (20C2)(0.05^17 + 0.8^3)^2 (1 - (0.05^17 + 0.8^3))^18
i don't know whether my ans is correct
2) 1400 unvaccinated children were exposed to whooping cough. What is the probability that at least 78% of these children develop infections?
My solution: this must be approx. to normal
P(Z >= 1092 - 1120 / square_root(224)) = p(Z < 1.87) = 0.9693
Please check my answers and let me know. Thanks in advance.
Let $V$ count the number of vaccinated children who develop whooping cough, and $U$ count the number of unvaccinated children who do so. These can be assumed to be independent and binomially distributed variables.
We have been given: $V\sim\mathcal{B}(17,0.05), U\sim\mathcal{B}(3,0.80), V\bot U$
We are tasked to find: $\mathrm{P}(V+U=2)$
We use the following: $$V\sim\mathcal{B}(17,0.05)\iff \mathrm{P}(V\!=\!v)={17\choose v}(0.05)^v(0.95)^{17-v}$$
$$U\sim\mathcal{B}(3,0.80)\iff \mathrm{P}(U\!=\!u)={3\choose u}(0.80)^u(0.20)^{3-u}$$
$$\mathrm{P}(V\!+\!U\!=\!2) = \mathrm{P}(V\!=\!0)\mathrm{P}(U\!=\!2)+\mathrm{P}(V\!=\!1)\mathrm{P}(U\!=\!1)+\mathrm{P}(V\!=\!2)\mathrm{P}(U\!=\!0)$$
Put it together.