Respuesta :
Answer:
(i) 0.46, (ii)0.42, and (iii)0.143
Step-by-step explanation:
Let [tex]p[/tex] be the probability of hitting the target and [tex]q[/tex] be the probability of missing the target by Laura.
Given that [tex]p=0.4\;\cdots (1)[/tex]
As Laura either hit or miss the target, so [tex]p+q=1[/tex].
[tex]\Rightarrow q=1-p=0.6 \; \cdots (2)[/tex]
Again, let [tex]r[/tex] be the probability of hitting the target and [tex]s[/tex] be the probability of missing the target by Philip.
Here, [tex]r=0.1\;\cdots (3)[/tex]
Similarly, [tex]\Rightarrow s=1-r=0.9\;\cdots (4)[/tex]
(i)The probability that the target is hit means that the target is not missed by both, either of one or both hit the target.
=1-(probability of missing the target by both)
=1-[tex]qr[/tex] [from equation (2) and (4) ]
=[tex]1-0.6\times 0.9[/tex]
=[tex]1-0.54[/tex]
=0.46
(ii) The probability that the target is hit by exactly one shot means either of one hit the target.
=Hit by Laura and missed by Philip or hit by Philip and missed by Laura
[tex]=0.4\times 0.9+0.1\times 0.6[/tex] [from equations (1),(4) and (3),(2)]
[tex]=0.36+0.06[/tex]
=0.42
(iii)Given that the target was hit by exactly one shot, so, the given probability is 0.42 [from (ii) part]
No, the probability that the target was hit by Philip = probability of hitting the target by Philip and missing the target by Laura
[tex]=0.1\times 0.6[/tex] [from equations (3) and (2)]
=0.06
So, the probability of hitting the target by Philip
[tex]=\frac {0.06}{0.42}[/tex]
[tex]=\frac {1}{7}[/tex]
[tex]=0.143[/tex] (approx)
