Question

Question : A player is holding a desert eagle and facing an enemy. Each shot he fires has 50% chance of hitting the enemy. If the shot is accurate there's also a 20% chance of it being a headshot which will kill the target immediately, otherwise it requires 3 shots to take down the enemy. What is the probability that the player take down the enemy in 7 shots?

Please help me out with this probability question.​

Answer 1

Given : A player is holding a desert eagle and facing an enemy.

Each shot he fires has 50% chance of hitting the enemy.

If the shot is accurate there's also a 20% chance of it being a headshot which will kill the target immediately,

otherwise it requires 3 shots to take down the enemy.

To Find : What is the probability that the player take down the enemy in 7 shots?

Solution:

Each shot he fires has 50% chance of hitting the enemy

= 0.5

. If the shot is accurate there's also a 20% chance of it being a headshot

Head shot = 0.5 * 0.2 = 0.1

Not head shot = 0.5 * 0.8 = 0.4

Not correct shot = 0.5

probability that the player take down the enemy in 7 shots

considering  exactly 7 shots   ( not with in  )

Case 1 :    7th shot is Head shot  

Head shot   =  0.1

Then there can be 0 , 1 or 2  not head shot in 1st 6

⁶C₀(0.4)⁰(0.5)⁶  + ⁶C₁(0.4)¹(0.5)⁵  +  ⁶C₂(0.4)²(0.5)⁴

=  (0.5)⁴ ( 0.5²  + 1.2 + 2.4)

=  (0.5)⁴ ( 3.85)

Probability = 0.1 * (0.5)⁴ ( 3.85)

= (0.5)⁴ (0.385)

Case 2 :    7th shot is not Head shot  

not Head shot   =  0.4

Then must be 2 head shot in 1st 6

 ⁶C₂(0.4)²(0.5)⁴

=   (2.4)(0.5)⁴

Probability  =   (2.4)(0.5)⁴(0.4)

=  (0.5)⁴(0.96)

Total Probability =  (0.5)⁴ (0.385) + (0.5)⁴(0.96)  

=  (0.5)⁴ (1.345)

= 0.0840625

= 8.40625 %

probability that the player take down the enemy in exactly 7 shots =  8.40625 %  or 0.0840625

If its with in 7 shots then Similarly

Calculate for 1 shot , 2 shots , 3 shots , 4 shots , 5 shots and 6 shots as well

and add

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