The probability that a bulb produced by a factory will use after 12 days of used in 0.05 .find the probability that out of 5 such bulbs(a)none(b)more than one will fuse after 150 days.
Answers
Answer:
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
Step-by-step explanation:
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hi dear
your answer is
option A
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)=nCxqn−xpx, where x=1,2,...n
=5Cx(0.95)5−x⋅(0.05
(i) P(none)=P(X=0)
=5C0(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)
=5C0(0.95)5×(0.05)0+5C1(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
(iv) P(atleastone)=P(X≥1)
=1−P(X<1)
=1−P(X=0)
=1−5C0(0.95)5×(0.05)0
=1−1×(0.95)5
=1−(0.95)5
hopes its helps you