Question

A rifle man firing at a distant target and has only 10% chance of hitting it. The minimum number of rounds he must fire in order to have 50% chance of hitting it atleast once is:-

(a) 6
(b) 7
(c) 8
(d) 9

Show proper steps :)​

Answer 1

Answer:

Given that the probability of hitting in one shot = 10/100 = 1/10

Therefore, the probability of not hitting = 1-(1/10) = 9/10

The probability of hitting at least once in “n” shots = 1- (9/10)n

From the given condition, 1- (9/10)n = ½

(9/10)n = ½

Taking logarithm on both the sides,we get n= 6.5 approximately.

Hence, the value of n is 7.

Answer 2

★ Concept :-

Here the concept of Probability and logarithmic series has been used. Firstly we can take the probability as given by removing percentage sign. Then from first case we can take the probability inverse. Then even for second case we can apply it. Then we shall apply logarithm there and thus find the answer.

Let's do it !!

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★ Formula Used:-

$\;\boxed{\sf{\pink{P(not\:E)\;=\;1\:-\:P(E)}}}$

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★ Solution :-

Given,

» Chance of hitting the target in one fire = P(E) = 10%

» Increased Probability = 50%

• Probability of hitting the target in one fire = 10/100 = ⅒ = P(E)

• Increased Probability = 50/100 = ½

Firstly we shall calculate the probability of not hitting the target in one fire. So let this be P(not E).

We have the formula as,

$\;\sf{\rightarrow\;\;P(not\:E)\;=\;1\:-\:P(E)}$

By applying values, we get

$\;\sf{\rightarrow\;\;P(not\:E)\;=\;1\:-\:\dfrac{1}{10}}$

$\;\sf{\rightarrow\;\;P(not\:E)\;=\;\dfrac{10\:-\:1}{10}}$

$\;\sf{\rightarrow\;\;P(not\:E)\;=\;\dfrac{9}{10}}$

This is the probability of not hitting in one chance. This means this will be probability of hitting in more than one chance.

Now let's calculate it for nth probability because we need to calculate the minimum number of chance required.

Thus the minimum number of chance = n

Now the probability of hitting at least once will be given as ,

$\;\;\tt{\mapsto\;\;P(not\:E)\;=\;1\;-\;P(E)}$

• Here the P(E) and P(not E) are different from above. They are just general notations for determining probability of an event.

By applying values, we get

$\;\;\tt{\mapsto\;\;P(not\:E)\;=\;1\;-\bigg[\dfrac{9}{10}\bigg]^{n}}$

• Here n is the minimum number of chances.

• Here we used 9/10 as Probability because it is probability of not hitting in one chance.

• Also the probability of hitting in many chance (P not E) as given in qúestion is 50% . So applying it, we shall get

$\;\;\tt{\mapsto\;\;P(not\:E)\;=\;1\;-\bigg[\dfrac{9}{10}\bigg]^{n}\;=\;\dfrac{1}{2}}$

$\;\;\tt{\mapsto\;\;P(not\:E)\;=\;\bigg[\dfrac{9}{10}\bigg]^{n}\;=\;1\;-\;\dfrac{1}{2}}$

$\;\;\tt{\mapsto\;\;P(not\:E)\;=\;\bigg[\dfrac{9}{10}\bigg]^{n}\;=\;\dfrac{2\;-\;1}{2}}$

$\;\;\tt{\mapsto\;\;P(not\:E)\;=\;\bigg[\dfrac{9}{10}\bigg]^{n}\;=\;\dfrac{1}{2}}$

Now we see that here comes the role of logarithm. Let's firstly solve the equation, here.

This can be written as,

$\;\;\tt{\leadsto\;\;n(log 9\:-\:log 10\;=\;log 1 \:-\:log 2)}$

This can be converted as,

$\;\;\tt{\leadsto\;\;n(2log 3\:-\:1)\;=\;-log 2}$

$\;\;\tt{\leadsto\;\;n\;=\;\dfrac{-log 2}{-(1\:-\:2log 3)}}$

$\;\;\tt{\leadsto\;\;n\;=\;\dfrac{log 2}{1\:-\:2log 3}}$

By changing the sign, we got it.

• log 2 = 0.3020

• log 3 = 0.0452

Here the base of log is 10.

So by applying the values, we get

$\;\;\tt{\leadsto\;\;n\;=\;\dfrac{0.03020}{1\:-\:2(0.0452)}}$

$\;\;\tt{\leadsto\;\;n\;=\;\dfrac{0.03020}{1\:-\:0.9542}}$

$\;\;\tt{\leadsto\;\;n\;=\;\dfrac{0.03020}{0.0452}}$

By using log functions,

$\;\;\bf{\leadsto\;\;n\;=\;6.68}$

Thus we got the value of 6.68

This can be rounded off as 6 times or 7 times since number of chances cannot be in decimals.

We see that if the number of chances is 6, then the probability comes out to be 53%.

And if the number of chances is 7, then the probability comes out to be 48%.

We see that 48 % is much closer to the given 50% than 53% .

So number of chances will be 7.

>> Minimum nunber of chances = 7