Question

Admitimos que o nascimento de meninas e meninos fosse igual. Um jovem

casal pretende ter 4 filhos, calcule a probabilidade de nascerem:

A) duas meninas;

b) pelo menos duas meninas;

c) no maximo duas meninas;

d) quatro meninas.

Answer 1

Explicação passo-a-passo:

a)

$\sf P=\dbinom{4}{2}\cdot\Big(\dfrac{1}{2}\Big)^2\cdot\Big(\dfrac{1}{2}\Big)^2$

$\sf P=6\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}$

$\sf P=\dfrac{6}{16}$

$\sf P=\dfrac{3}{8}$

$\sf \red{P=37,5\%}$

b)

• Duas meninas

$\sf P=\dbinom{4}{2}\cdot\Big(\dfrac{1}{2}\Big)^2\cdot\Big(\dfrac{1}{2}\Big)^2$

$\sf P=6\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}$

$\sf P=\dfrac{6}{16}$

$\sf P=\dfrac{3}{8}$

$\sf \red{P=37,5\%}$

• Três meninas

$\sf P=\dbinom{4}{3}\cdot\Big(\dfrac{1}{2}\Big)^3\cdot\Big(\dfrac{1}{2}\Big)^1$

$\sf P=4\cdot\dfrac{1}{8}\cdot\dfrac{1}{2}$

$\sf P=\dfrac{4}{16}$

$\sf P=\dfrac{1}{4}$

$\sf \red{P=25\%}$

• Quatro meninas

$\sf P=\dbinom{4}{4}\cdot\Big(\dfrac{1}{2}\Big)^4$

$\sf P=1\cdot\dfrac{1}{16}$

$\sf P=\dfrac{1}{16}$

$\sf \red{P=6,25\%}$

A probabilidade de nascerem pelo menos duas meninas é:

$\sf P=\dfrac{3}{8}+\dfrac{1}{4}+\dfrac{1}{16}$

$\sf P=\dfrac{6+4+1}{16}$

$\sf P=\dfrac{11}{16}$

$\sf \red{P=68,75\%}$

c)

• Nenhuma menina

$\sf P=\dbinom{4}{0}\cdot\Big(\dfrac{1}{2}\Big)^4$

$\sf P=1\cdot\dfrac{1}{16}$

$\sf P=\dfrac{1}{16}$

$\sf \red{P=6,25\%}$

• Uma menina

$\sf P=\dbinom{4}{1}\cdot\Big(\dfrac{1}{2}\Big)^1\cdot\Big(\dfrac{1}{2}\Big)^3$

$\sf P=4\cdot\dfrac{1}{2}\cdot\dfrac{1}{8}$

$\sf P=\dfrac{4}{16}$

$\sf P=\dfrac{1}{4}$

$\sf \red{P=25\%}$

• Duas meninas

$\sf P=\dbinom{4}{2}\cdot\Big(\dfrac{1}{2}\Big)^2\cdot\Big(\dfrac{1}{2}\Big)^2$

$\sf P=6\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}$

$\sf P=\dfrac{6}{16}$

$\sf P=\dfrac{3}{8}$

$\sf \red{P=37,5\%}$

A probabilidade de nascerem no máximo duas meninas é:

$\sf P=\dfrac{1}{16}+\dfrac{1}{4}+\dfrac{3}{8}$

$\sf P=\dfrac{1+4+6}{16}$

$\sf P=\dfrac{11}{16}$

$\sf \red{P=68,75\%}$

d)

$\sf P=\dbinom{4}{4}\cdot\Big(\dfrac{1}{2}\Big)^4$

$\sf P=1\cdot\dfrac{1}{16}$

$\sf P=\dfrac{1}{16}$

$\sf \red{P=6,25\%}$