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 = 8Probability (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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