Question

the probability of a bomb fitting the bridges 1 by 2 and 3 direct fix are needed to destroy it not necessarily consecutive find the minimum number of bombs required so the probability of the bridge being destroyed is greater than 0.9 ​

Answer 1

Answer:

Let n be the least number of bombs required and X the number of bombs that hit the bridge.

Then X follows a binomial distribution with parameters n and p=

2

1

.

⇒q=1−p=

2

1

Now P(X≥2)>0.9⇒1−P(X<2)>0.9

⇒P(X=0)+P(X=1)<0.1

⇒

n

C

0

(

2

1

)

n

+

n

C

1

(

2

1

)

n−1

(

2

1

)<

10

1

⇒

2

n

n+1

<

10

1

⇒10(n+1)<2

n

By trial and error, we get n≥7.

Thus, the least value of n is 7.