Question 7, 4.2.21
HW Score: $40.28\mathrm{\%},4.83$ of 12
Part 3 of 3
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$63\mathrm{\%}$ of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four:
(a) $P(5)=0.173$ (Round to three decimal places as needed.)
(b) $P(x\ge 6)=0.706$ (Round to three decimal places as needed.)
(c) $P(x<4)=◻($ Round to three decimal places as needed.)

Step 1 ：The formula for binomial probability is: $P(x)=C(n,x)\ast (p^x)\ast ((1-p{)}^{n-x})$

Step 2 ：For part (c), we need to find $P(x<4)$, which is the sum of $P(0)$, $P(1)$, $P(2)$, and $P(3)$.

Step 3 ：Calculate $P(0)=C(10,0)\ast ({0.63}^0)\ast ((1-0.63{)}^{10-0})=1\ast (1)\ast ({0.37}^{10})=0.00014$

Step 4 ：Calculate $P(1)=C(10,1)\ast ({0.63}^1)\ast ((1-0.63{)}^{10-1})=10\ast (0.63)\ast ({0.37}^9)=0.00137$

Step 5 ：Calculate $P(2)=C(10,2)\ast ({0.63}^2)\ast ((1-0.63{)}^{10-2})=45\ast ({0.63}^2)\ast ({0.37}^8)=0.00675$

Step 6 ：Calculate $P(3)=C(10,3)\ast ({0.63}^3)\ast ((1-0.63{)}^{10-3})=120\ast ({0.63}^3)\ast ({0.37}^7)=0.02077$

Step 7 ：Adding these probabilities together will give us $P(x<4)$.

Step 8 ：Calculate $P(x<4)=P(0)+P(1)+P(2)+P(3)=0.00014+0.00137+0.00675+0.02077=0.02903$

Step 9 ：$\boxed{{\displaystyle P(x<4)=0.02903}}$, so the probability that less than four U.S. adults have very little confidence in newspapers is approximately 0.029 or 2.9%.