a lot size 10% defective item 10 items are chosen randomly from this lot the probability that
at least two of chosen items are defective its
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Answer:
Total no. of items =100
No. of defective items =10
No. of safe items =100−10=90
Let p and q be the probability of defective and safe items respectively.
p=
100
10
=
10
1
q=
100
90
=
10
9
n=5
As we know that,
P(x=r)=
n
C
r
(p)
r
(q)
n−r
Therefore,
P(x≤1)=P(x=0)+P(x=1)
P(x≤1)=
5
C
0
(
10
1
)
0
(
10
9
)
5
+
5
C
1
(
10
1
)
1
(
10
9
)
4
P(x≤1)=(
10
9
)
4
(
10
9
+
10
5
)
⇒P(x≤1)=(
10
9
)
4
(
5
7
)
Hence the probability that the store will receive at most one defective item is (
10
9
)
4
(
5
7
).
Hence the correct answer is (
10
9
)
4
(
5
7
).
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