Question

probability that a bulb fuse within a year is 0.05 . find the probability out of 5 bulb 3 bulb will fuse within a year​

Answer 1

Step-by-step explanation:

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

It is given that, p=0.05

∴q=1−p=1−0.05=0.95

X has a binomial distribution with n=5 and p=0.05

∴P(X=x)=

n

C

x

q

n−x

p

x

, where x=1,2,...n

=

5

C

x

(0.95)

5−x

⋅(0.05)

x

(i) P(none)=P(X=0)

=

5

C

0

(0.95)

5

⋅(0.05)

0

=1×(0.95)

5

=(0.95)

5

(ii) P(notmorethanone)=P(X≤1)

=P(X=0)+P(X=1)

=

5

C

0

(0.95)

5

×(0.05)

0

+

5

C

1

(0.95)

4

×(0.05)

1

=1×(0.95)

5

+5×(0.95)

4

×(0.05)

=(0.95)

5

+(0.25)(0.95)

4

=(0.95)

4

[0.95+0.25]

=(0.95)

4

×1.2

(iii) P(morethan1)=P(X>1)

=1−P(X≤1)

=1−P(notmorethan1)

=1−(0.95)

4

×1.2

(iv) P(atleastone)=P(X≥1)

=1−P(X<1)

=1−P(X=0)

=1−

5

C

0

(0.95)

5

×(0.05)

0

=1−1×(0.95)

5

=1−(0.95)

5