Explanation
He will hit the target, $$\displaystyle P(A)=\dfrac{1}{5}.$$ He will not hit the target,$$\displaystyle P\bar{(A)}=\dfrac{4}{5}.$$ Probability that he will not hit the target in 10 shoots is $$\displaystyle \left ( \dfrac{4}{5} \right )^{10}$$ So, at least once, target will be hit.Required probability $$\displaystyle =1-\left ( \dfrac{4}{5} \right )^{10}$$
Let the coin be tossed $$n$$ times.
Let getting head is consider to be success.$$\therefore p=\dfrac{1}{2}, q=1-p=1-\dfrac{1}{2}=\dfrac{1}{2}$$It is given that, $$p(X=3)=p(x=5)$$
$$\Rightarrow {}^nC_3\left(\dfrac{1}{2}\right)^3\left(\dfrac{1}{2}\right)^{n-3}={}^nC_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{n-5}$$$$\Rightarrow ^nC_3=^nC_5$$
$$\Rightarrow n=3+5$$ $$[\because ^nC_x=^nC_y\Rightarrow x+y=n]$$$$\Rightarrow n=8$$
Now, $$P(x=1)=\ ^8C_1\left(\dfrac{1}{2}\right)^1\left(\dfrac{1}{2}\right)^{8-1}$$
$$=^8C_1\times \left(\dfrac{1}{2}\right)^8=\dfrac{1}{32}$$
We have, $$p = \dfrac{1}{2}, q = \dfrac{1}{2}, n = 8$$Probability (at least 6 heads) $$= {}^8C_6\times \left(\dfrac{1}{2}\right)^6\times \left(\dfrac{1}{2}\right)^2 + {}^8C_7 \times\left(\dfrac{1}{2}\right)^{7} \left(\dfrac{1}{2}\right)+{}^{8}C_{8}\left(\dfrac{1}{2}\right)^8$$$$=\left(\dfrac{8\times7}{1\times2}+8+1\right) \left(\dfrac{1}{2}\right)^8$$$$= 37 \times \dfrac{1}{256} = \dfrac{37}{256}$$
Given, $$9P(X=4)=P(X=2)$$
$$\therefore 9\ ^6C_4\ p^4(1-p)^2=\ ^6C_2\ p^2(1-p)^4$$
$$\Rightarrow \dfrac{(1-p)^2}{p^2}=9$$ ............ $$\because \ ^6C_2=^6C_4$$
$$\Rightarrow \dfrac{1-p}{p}=3$$
$$\Rightarrow p=\dfrac {1}{4}$$
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