Bomb The Target! Part 3

Probability Level 5

A Lockheed Martin F-22 Raptor has gone on a mission to bomb an old military hideout . Three bombs are dropped in succession at it .

The probabilities of a hit in the first shot is 1 2 \frac{1}{2} , in the second is 2 3 \frac{2}{3} and in the third is 3 4 \frac{3}{4} .

In case of one hit , the probability of destroying the target is 2 3 \frac{2}{3} , in case of two hits is 7 11 \frac{7}{11} and in case of three hits is 1 1 .

Find the probability of destroying the hideout in three shots .

The answer is of the form f g \frac{f}{g} where f f and g g are co-prime numbers ;

Report the last three digits of f g f^{g} .

You can try more of my Questions here .


The answer is 121.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Alex Li
May 21, 2015

Wait, so the probability of destroying it with 2 hits is less than that with 1 hit? That's kind of unusual.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

P (hideout is destroyed) = P (exactly 1 hit) ⋅ P (hideout is destroyed) (exactly 1 hit) + P (exactly 2 hits) ⋅ P (hideout is destroyed) (exactly 2 hits) + P (exactly 3 hits) ⋅ P (hideout is destroyed) (exactly 3 hits) P \text{ (hideout is destroyed) } =P \text{ (exactly 1 hit) } \cdot P\frac{ \text{ (hideout is destroyed) } }{ \text{ (exactly 1 hit) } } \\ +P \text{ (exactly 2 hits) } \cdot P \frac{ \text{ (hideout is destroyed) } }{ \text{ (exactly 2 hits) } } \\ + P \text{ (exactly 3 hits) } \cdot P \frac{ \text{ (hideout is destroyed) } }{ \text{ (exactly 3 hits) } }

Now,

P (exactly 1 hit) = P ( H ) ⋅ P ( M ) ⋅ P ( M ) + P ( M ) ⋅ P ( H ) ⋅ P ( M ) + P ( M ) ⋅ P ( M ) ⋅ P ( H ) = ( 1 2 ⋅ 1 3 ⋅ 1 4 ) + ( 1 2 ⋅ 2 3 ⋅ 1 4 ) + ( 1 2 ⋅ 1 3 ⋅ 3 4 ) = ( 1 4 ) P \text{ (exactly 1 hit) } = P(H) \cdot P(M) \cdot P(M) + P(M) \cdot P(H) \cdot P(M) \\+ P(M) \cdot P(M) \cdot P(H) \\= ( \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} ) + ( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} ) + ( \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} ) = ( \frac{1}{4} )

Now,

P (exactly 2 hits) = P ( H ) ⋅ P ( H ) ⋅ P ( M ) + P ( H ) ⋅ P ( M ) ⋅ P ( H ) + P ( M ) ⋅ P ( H ) ⋅ P ( H ) = ( 1 2 ⋅ 2 3 ⋅ 1 4 ) + ( 1 2 ⋅ 1 3 ⋅ 3 4 ) + ( 1 2 ⋅ 2 3 ⋅ 3 4 ) = ( 11 24 ) P \text{ (exactly 2 hits) } = P(H) \cdot P(H) \cdot P(M) + P(H) \cdot P(M) \cdot P(H) \\+ P(M) \cdot P(H) \cdot P(H) \\=( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} ) + ( \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} ) + ( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} ) = ( \frac{11}{24} )

Now,

P (exactly 3 hits) = P ( H ) ⋅ P ( H ) ⋅ P ( H ) = 1 2 ⋅ 2 3 ⋅ 3 4 = 1 4 P \text{ (exactly 3 hits) } = P(H) \cdot P(H) \cdot P(H) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}

P (destroying the hideout) = ( 1 4 ⋅ 2 3 ) + ( 11 24 ⋅ 7 11 ) + ( 1 4 ⋅ 1 ) = 17 24 P \text { (destroying the hideout) } = ( \frac{1}{4} \cdot \frac{2}{3} ) + ( \frac{11}{24} \cdot \frac{7}{11} ) + ( \frac{1}{4} \cdot 1) = \frac{17}{24}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

P (hideout is destroyed) = P (exactly 1 hit) ⋅ P (hideout is destroyed) (exactly 1 hit) + P (exactly 2 hits) ⋅ P (hideout is destroyed) (exactly 2 hits) + P (exactly 3 hits) ⋅ P (hideout is destroyed) (exactly 3 hits) P \text{ (hideout is destroyed) } =P \text{ (exactly 1 hit) } \cdot P\frac{ \text{ (hideout is destroyed) } }{ \text{ (exactly 1 hit) } } \\ +P \text{ (exactly 2 hits) } \cdot P \frac{ \text{ (hideout is destroyed) } }{ \text{ (exactly 2 hits) } } \\ + P \text{ (exactly 3 hits) } \cdot P \frac{ \text{ (hideout is destroyed) } }{ \text{ (exactly 3 hits) } }

Now,

P (exactly 1 hit) = P ( H ) ⋅ P ( M ) ⋅ P ( M ) + P ( M ) ⋅ P ( H ) ⋅ P ( M ) + P ( M ) ⋅ P ( M ) ⋅ P ( H ) = ( 1 2 ⋅ 1 3 ⋅ 1 4 ) + ( 1 2 ⋅ 2 3 ⋅ 1 4 ) + ( 1 2 ⋅ 1 3 ⋅ 3 4 ) = ( 1 4 ) P \text{ (exactly 1 hit) } = P(H) \cdot P(M) \cdot P(M) + P(M) \cdot P(H) \cdot P(M) \\+ P(M) \cdot P(M) \cdot P(H) \\= ( \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} ) + ( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} ) + ( \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} ) = ( \frac{1}{4} )

Now,

P (exactly 2 hits) = P ( H ) ⋅ P ( H ) ⋅ P ( M ) + P ( H ) ⋅ P ( M ) ⋅ P ( H ) + P ( M ) ⋅ P ( H ) ⋅ P ( H ) = ( 1 2 ⋅ 2 3 ⋅ 1 4 ) + ( 1 2 ⋅ 1 3 ⋅ 3 4 ) + ( 1 2 ⋅ 2 3 ⋅ 3 4 ) = ( 11 24 ) P \text{ (exactly 2 hits) } = P(H) \cdot P(H) \cdot P(M) + P(H) \cdot P(M) \cdot P(H) \\+ P(M) \cdot P(H) \cdot P(H) \\=( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} ) + ( \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} ) + ( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} ) = ( \frac{11}{24} )

Now,

P (exactly 3 hits) = P ( H ) ⋅ P ( H ) ⋅ P ( H ) = 1 2 ⋅ 2 3 ⋅ 3 4 = 1 4 P \text{ (exactly 3 hits) } = P(H) \cdot P(H) \cdot P(H) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}

P (destroying the hideout) = ( 1 4 ⋅ 2 3 ) + ( 11 24 ⋅ 7 11 ) + ( 1 4 ⋅ 1 ) = 17 24 P \text { (destroying the hideout) } = ( \frac{1}{4} \cdot \frac{2}{3} ) + ( \frac{11}{24} \cdot \frac{7}{11} ) + ( \frac{1}{4} \cdot 1) = \frac{17}{24}

Hi Rajorshi , can you please copy the LaTex from my comment and repost your solution . Sorry for being intrusive but I believe it looks better now .

BTW nice solution , the way that I had used was a bit longer and it certainly involved a lot of LaTex , so I believe your solution is enough .

@Rajorshi Chaudhuri

A Former Brilliant Member - 6 years, 4 months ago

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Thanks a lot Azhaghu for helping out with the LaTeX \LaTeX . It looks quite neat now.

Rajorshi Chaudhuri - 6 years, 4 months ago

@Azhaghu Roopesh M Nice to see you helping others with L A T E X LATEX , but it would be better if you'll use text without L A T E X LATEX . Though you have used 'text' markdown, but that's better outside \(

Pranjal Jain - 6 years, 3 months ago

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Thanks Pranjal ⌣ ¨ \ddot\smile

Sorry for the late response since I have not been checking my e-mail lately .

And sorry about that , you see my LaTeX \LaTeX knowledge is limited , just knowing the basics ,all I can do is correct the LaTeX \LaTeX here and there so that the solution looks good .

A Former Brilliant Member - 6 years, 3 months ago

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