Shooters

In the army, regiments are six shooters. The first shooter target hit with a probability of 49%, next with 75%, 41%, 20%, 34%, 63%. Calculate the probability of target hit when shooting all at once.

Correct answer:

p =  98.53 %

Step-by-step explanation:

p1=0.49+0.750.490.75=0.873 p2=p1+0.41p10.41=0.925 p3=p2+0.2p20.2=0.94 p4=p3+0.34p30.34=0.96 p5=p4+0.63p40.63=0.985  p=100p598.53%



Did you find an error or inaccuracy? Feel free to write us. Thank you!



Showing 1 comment:
Chris Roberts
Let's denote:

H: Hit
M: Miss
Now, we can represent the probabilities as follows:

Shooter 1: P(H1) = 0.49, P(M1) = 1 - P(H1) = 0.51
Shooter 2: P(H2) = 0.75, P(M2) = 1 - P(H2) = 0.25
Shooter 3: P(H3) = 0.41, P(M3) = 1 - P(H3) = 0.59
Shooter 4: P(H4) = 0.20, P(M4) = 1 - P(H4) = 0.80
Shooter 5: P(H5) = 0.34, P(M5) = 1 - P(H5) = 0.66
Shooter 6: P(H6) = 0.63, P(M6) = 1 - P(H6) = 0.37
Now, we want to find the probability of all shooters hitting the target when shooting at once. We can use the multiplication rule for independent events, which states that the probability of all independent events occurring together is the product of their individual probabilities.

Let's calculate the probability of hitting the target when shooting all six shooters at once:

P(all shooters hit) = P(H1) * P(H2) * P(H3) * P(H4) * P(H5) * P(H6)
P(all shooters hit) = 0.49 * 0.75 * 0.41 * 0.20 * 0.34 * 0.63
P(all shooters hit) ≈ 0.008728 (rounded to six decimal places)

So, the probability of hitting the target when shooting all six shooters at once is approximately 0.008728 or about 0.873%.





Tips for related online calculators
Would you like to compute the count of combinations?

You need to know the following knowledge to solve this word math problem:

Related math problems and questions: