Question

In 1995, 70% of all children in the U.S. were living with both parents. If 25 children were selected at random in the U.S., what is the probability that at most 10 of them will be living with both of their parents? Round your answer to 4 decimal places.

Answer 1

Answer: 0.0018

Step-by-step explanation:

Binomial distribution formula :-

$P(x)=^nC_xp^x(q)^{n-x}$, here P(x) is the probability of getting success at x trial , n is the total number of trails, p is the probability of getting success in each trail.

Given : The probability that a child in the U.S. was living with both parents : p=0.70 ; q=1-0.70=0.30

If 25 children were selected at random in the U.S.,then the probability that at most 10 of them will be living with both of their parents will be :-

$P(x\leq10)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)\\\\=^{25}C_{0}(0.7)^0(0.3)^{25}+^{25}C_{1}(0.7)^1(0.3)^{24}+^{25}C_{2}(0.7)^2(0.3)^{23}+^{25}C_{3}(0.7)^3(0.3)^{22}+^{25}C_{4}(0.7)^4(0.3)^{21}+^{25}C_{5}(0.7)^5(0.3)^{20}+^{25}C_{6}(0.7)^6(0.3)^{19}+^{25}C_{7}(0.7)^7(0.3)^{`18}+^{25}C_{8}(0.7)^8(0.3)^{17}+^{25}C_{9}(0.7)^9(0.3)^{16}+^{25}C_{10}(0.7)^{10}(0.3)^{15}\\\\=(0.7)^0(0.3)^{25}+25(0.7)^1(0.3)^{24}+300(0.7)^2(0.3)^{23}+2300(0.7)^3(0.3)^{22}+ 12650(0.7)^4(0.3)^{21}+53130(0.7)^5(0.3)^{20}+177100(0.7)^6(0.3)^{19}+480700(0.7)^7(0.3)^{`18}+1081575(0.7)^8(0.3)^{17}+2042975(0.7)^9(0.3)^{16}+3268760(0.7)^{10}(0.3)^{15}\\\\=0.00177840487034\approx0.0018$

Hence, the probability that at most 10 of them will be living with both of their parents is 0.0018.