Question

The probability of bomb hitting a bridge is 1/2 and two direct hits
are required to destroy it.The least number of bombs
required so that the probability of the bridge being destroyed
is greater than 0.9 is

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{The probability of bomb hitting a bridge is}\ \dfrac12\ \text{and two direct hits} \hspace{.33em}\\~\\
& \normalsize \text{ are required to destroy it.The least number of bombs}\hspace{.33em}\\~\\
& \normalsize \text{ required so that the probability of the bridge being destroyed}\hspace{.33em}\\~\\
& \normalsize \text{ is greater than 0.9 is }\hspace{.33em}\\~\\
& a.)\ 7\ \ bombs \hspace{.33em}\\~\\
& b.)\ 3\ \ bombs \hspace{.33em}\\~\\
& c.)\ 8\ \ bombs \hspace{.33em}\\~\\
& d.)\ 9\ \ bombs \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

is this a poisson distribution ?

Answer 3

i found a page i cant understand soln here

http://mathhelpforum.com/statistics/81015-probability-problems.html

Answer 4

ok, or binomial

Answer 5

i hvent studied poision ditributuion , i have this in probability chapter

Answer 6

then binomial looks appropriate

Answer 7

i also didnt studied binomial distributuion

Answer 8

what techniques are in this chapter ?

Answer 9

Use of AND

use of OR

Combination of AND and OR

Random experiment

sample space

Event

Non-event

Imposiible event

mutually exclusive event

independent event

conditional probablity

Answer 10

I guess we can use all of that to derive the binomial distribution, and then solve the problem.

For example, what is the chances we have 0 hits in n tries ?
Can you do that problem ?

Answer 11

1/2

Answer 12

it is 1/2 if we tried once
but what if we try n times ?

Answer 13

(1/2)^n

Answer 14

yes.

next, what are the chances we get exactly 1 hit in n tries ?

Answer 15

(1/2)^n

Answer 16

First, what is the prob of hitting on the first shot, and missing on all the others ?
1/2 * 1/2^(n-1) = 1/2^n

but we could miss the first shot, hit on the second, and miss on the rest. that is also 1/2^n
and so on...

Answer 17

in other words, the chance of making exactly one hit is n * (1/2)^n
ok ?

Answer 18

ok

Answer 19

so we know the chances of getting exactly 0 or 1 hit
1 - that prob is the chance of getting 2 (or more) hits in n tries.

Answer 20

in other words,
Pr(2 or more hits) = \( 1 - \frac{1}{2^n} - \frac{n}{2^n} \)
we want that to be bigger than 0.9
this is not easy to solve, except numerically, so use trial and error.

Answer 21

in other words, the chance of making exactly one hit is \(\color{red}{n}\) * (1/2)^n
ok ?

where does \(\color{red}{n}\) comes from

Answer 22

say we try 3 times
then we could have
x 0 0
0 x 0
0 0 x
each line has a 1/8 chance
and we have a total of 3/8 chance of getting exactly one hit.

Answer 23

thnks