A certain type of missile hits the target with probability p = 0.3. Wh...
Probability of hittiy the forget = 0.3
If'n' is the no. of times that the Missile is fired.
Probability of hitting at least once = 1-[1-0.3]
n = 0.8
0.7
n=0.2
n log 0.7 = log 0.2
⇒n=4.512
for n = 4 ; p < 0 . 8
taken = 5
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A certain type of missile hits the target with probability p = 0.3. Wh...
To solve this problem, we can use the concept of the binomial distribution. The binomial distribution is used to model situations where there are only two possible outcomes, in this case, hitting or missing the target.
Let's denote the probability of hitting the target as p = 0.3 and the probability of missing the target as q = 1 - p = 0.7.
We want to find the least number of missiles that should be fired so that there is at least an 80% probability of hitting the target.
Let's denote the number of missiles fired as n.
To find the probability of hitting the target at least once in n missiles, we can use the complement rule. The complement of hitting the target at least once is missing the target every time.
The probability of missing the target in one missile is q = 0.7. The probability of missing the target in n missiles is q^n.
So, the probability of hitting the target at least once in n missiles is 1 - q^n.
We want this probability to be at least 80%, so we have the inequality:
1 - q^n ≥ 0.8
Let's solve this inequality for n:
1 - 0.7^n ≥ 0.8
0.7^n ≤ 0.2
Taking the logarithm of both sides:
log(0.7^n) ≤ log(0.2)
n * log(0.7) ≤ log(0.2)
n ≥ log(0.2) / log(0.7)
Using a calculator, we find that n ≥ 4.33.
Since n must be a positive integer, the least number of missiles that should be fired is 5.
Therefore, the correct answer is option A) 5.