Question

A can hit the target 3 times in 6 shots, B 2 times in 6 shots
and C 4 times in 6 shots. They fire a volley. What is the
probablity that at least 2 shots hit ?

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{ A can hit the target 3 times in 6 shots, B 2 times in 6 shots }\hspace{.33em}\\~\\
& \normalsize \text{ and C 4 times in 6 shots. They fire a volley. What is the }\hspace{.33em}\\~\\
& \normalsize \text{ probablity that at least 2 shots hit ?}\hspace{.33em}\\~\\
& a.)\ \dfrac{1}{2} \hspace{.33em}\\~\\
& b.)\ \dfrac{1}{3} \hspace{.33em}\\~\\
& c.)\ \dfrac{2}{3} \hspace{.33em}\\~\\
& d.)\ \dfrac{3}{4} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

A can hit the target 3 times in 6 shots, B 2 times in 6 shots
and C 4 times in 6 shots. They fire a volley. What is the
probablity that at least 2 shots hit ?

a.)1/2
b.)1/3
c.)2/3
d.)3/4

Answer 3

\[p = P(A \cap B) +P(B \cap C) +P(A \cap C) +P(A \cap B \cap C) \]

Answer 4

i calculated 5/6 by ur formula

Answer 5

is it correct

Answer 6

So A, B and C will be fired all at once? Okay.

prob A can hit the target: 1/2

prob B:1/3

prob C: 2/3

If at least 2 targets have to be hit, we should calculate the combined probability of the following possibilities:

No targets are going to be hit.

\[\frac{ 1 }{ 2 } *\frac{ 2 }{ 3 } * \frac{ 1 }{ 3 } = \frac{ 1 }{ 9 }\]

1 target will be hit.

For this scenario, there are multiple permutations.

A hits, but not B/C - prob = 1/9

B hits, but not A/C - prob = 1/18

C hits, but not A/B - prob = 2/9

added together, you have a probability of: 7/18

Add to the probability that none of them hit.

7/18 + 1/9 = 9/18

The answer should be 1/2

Answer 7

I should add that you must subtract the probability "none of them hit" from 1 to get the right value, but since the value (according to my calculations) was 1/2, it would have been the same either way.

Answer 8

I think its 1/3 because 2/6 = 1/3

Answer 9

Jade's answer looks good
to review, using \(\bar{A}\) to mean A missed
Chance of hit and miss are as follows
A 1/2 \(\bar{A}\) 1/2
B 1/3 \(\bar{B}\) 2/3
C 2/3 \(\bar{C}\) 1/3

one way to do it is
Pr(2 hits or more) = 1 - Pr(0 hits) - Pr(1 hit)

To find Pr(0 hits) we multiply the chance each misses, namely
\(\bar{A}\bar{B}\bar{C}\) =1/2 * 2/3 * 1/3 = 2/18

To find the Pr(1 hit) we add up the prob of
\(AB\bar{C} + A\bar{B}C + \bar{A}BC= 1/2*1/3*1/3 + 1/2*2/3*2/3+1/2*1/3*2/3\)
which is 1/18+4/18+2/18= 7/18

so we have
Pr(2 hits or more) = 1 - Pr(0 hits) - Pr(1 hit)
Pr(2 hits or more) = 1 -2/18 - 7/18 = 1 - 1/2 = 1/2

Answer 10

yeah I ve made a stupid mistake, my equation has to be

\[\begin{align}
p &= P(A\cap B \cap \bar C) + P(A\cap C \cap \bar B) + P(B\cap C \cap \bar A) + P(A\cap B \cap C)\\
&=P(A)P(B)P(\bar C)+P(A)P(\bar B)P( C)+P(\bar A)P(B)P(C)+P(A)P(B)P( C) \\
&= \left(\frac 1 2 \times\frac 1 3\times \frac 1 3 \right)+\left(\frac 1 2 \times\frac 2 3\times \frac 2 3 \right)+\left(\frac 1 2 \times\frac 1 3\times \frac 2 3 \right)+\left(\frac 1 2 \times\frac 1 3\times \frac 2 3 \right) \\&=\frac 1 2
\end{align}\]

sorry guys