Question

The probability that a bulb produced by a factory will fuse after 150 days of used is 0.05. Find the probability that out of 5 such bulbs
(i) none

(ii) not more than one

(iii) more than one

(iv) atleast one will fuse after 150 days of use

Answer 1

Let X represents the number of bulbs that will fuse after 150 days of use in an experiment of 5 trails. The trials are Bernoulli trails.

p=P(success) = 0.05 and q = 1-p=1-0.05 = 0.95

X has a binomial distrubtion with n= 5, p=0.05 and q=0.95

P(X=r) = ${}^5{C}_r.\left(0.05{)}^r\right(0.95{)}^{5-r}$

(i) Required probability = P(X-r) = ${}^5{C}_0{p}^0{q}^5=q^5=(0.05{)}^5$

Let X represents the number of bulbs that will fuse after 150 days of use in an experiment of 5 trails. The trials are Bernoulli trails.

p=P(success) = 0.05 and q = 1-p=1-0.05 = 0.95

X has a binomial distrubtion with n= 5, p=0.05 and q=0.95

P(X=r) = ${}^5{C}_r.\left(0.05{)}^r\right(0.95{)}^{5-r}$
(ii) Required probability =$P(X\le 1)=P\left(0\right)+P\left(1\right)={}^5{C}_0{p}^0{q}^5=q^5={}^5{C}_1{p}^1{q}^4=q^5+5p{q}^4=q^4(q+5p)=\left(0.95{)}^4\right[0.95+5\times \left(0.05\right)]=(0.95{)}^4[0.95+0.25]=(0.95{)}^4\times 1.2$

Let X represents the number of bulbs that will fuse after 150 days of use in an experiment of 5 trails. The trials are Bernoulli trails.

p=P(success) = 0.05 and q = 1-p=1-0.05 = 0.95

X has a binomial distrubtion with n= 5, p=0.05 and q=0.95

P(X=r) = ${}^5{C}_r.\left(0.05{)}^r\right(0.95{)}^{5-r}$
(iii) Required probability =$P(X>1)=1-\left[P\right(0)+P(1\left)\right]=1-(0.95{)}^4\times 1.2$

Let X represents the number of bulbs that will fuse after 150 days of use in an experiment of 5 trails. The trials are Bernoulli trails.

p=P(success) = 0.05 and q = 1-p=1-0.05 = 0.95

X has a binomial distrubtion with n= 5, p=0.05 and q=0.95

P(X=r) = ${}^5{C}_r.\left(0.05{)}^r\right(0.95{)}^{5-r}$
(iv) Required probability = P(atleast one) = $=P(X\ge 1)=1-P\left(0\right)=1-(0.95{)}^5$