Answer:
a
[tex]P(2\leq X\leq 5)=0.875[/tex]
b
[tex]P(X<3)=0.3432[/tex]
Step-by-step explanation:
Let the random X variable representing the 6 companies that give 4 weeks of vacation after 15 years of employment:
-let p=0.5 be the probability of vacation. Since the companies are independent, X assumes a binomial random variable:[tex]P(X=x)=b(x;6,0.5)\\\\={6 \choose x}(0.5)^x(0.5)^{6-x}\\\\={6 \choose x}(0.5)^6, \ x=0,1,2,3...\ \ \ eqtn1[/tex]
#Probability that the number of companies that give vacation is anywhere from 2 to 5:
We use equation 1;
[tex]P(2\leq X\leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\={6 \choose 2}(0.5)^6+{6 \choose 3}(0.5)^6{6 \choose 4}(0.5)^6+{6 \choose 5}(0.5)^6\\\\=0.0156(15+20+15+6)\\\\=0.875[/tex]
Hence the probability that between 2 and 5 companies give vacation is 0.875
b. The probability that fewer than 3 companies give vacation is calculated as:
From equation one we get:
[tex]P(X<3)=P(X=0)+P(X=1)+P(X=2)\\\\={6 \choose 0}(0.5)^6+{6 \choose 1}(0.5)^6{6 \choose 2}(0.5)^6\\\\=0.0156(1+6+15)\\\\=0.3432[/tex]
Hence the probability that less than three companies give vacation is 0.3432
