A certain type of missile hits the target with probability p = 0.3. What is the least number of missiles should be fired, so that there is atleast an 80% probability that the target is hit?
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A certain type of missile hits the target with probability p = 0.3. What is the least number of missiles should be fired so that there is at least on 80% probability that the target is hit?
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NDA (Held On: 17 April 2016) Maths Previous Year paper
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5
6
7
None of the above
Answer (Detailed Solution Below)
Option 1 : 5
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Detailed Solution
Concept:
The binomial distribution formula is:
P (X = r) =
Where,
P(X) = probability of a success on an individual trial
p = probability of “successes”
r = number of success
q = probability of “failure” OR q = 1- p
n = number of trials
Calculation:
Given:
Probability of success, p = 0.3
To find: Number of trials for getting 80% probability (n) =?
Here, p = 0.3
So probability of missed target = q = 1 – 0.3 = 0.7
P (at least one) ≥ 0.8
Now, 1 – P(X = 0)
=
> 0.8
⇒ 1 - 1(0.7)n > 0.8
⇒ 1 - 0.8 > (0.7)n
⇒ 0.2 ≥ (0.7)n
Now, if n = 2, 0.2 ≥ 0.49 which is not correct.
If n = 4, 0.2 ≥ 0.24
∴ n = 5 as it gives 0.2 > 0.16 which is correct
Hence, option (3) is correct.( please follow )
Answer:
90° present of target and we can get our same answer
Step-by-step explanation:
step by step solution are same answer