Q.(a) During war, 1 ship out of 9 was sunk on an average in making a certain voyage. What was the probability that :

(a) Models the number of ships arriving safely as a binomial random variable with $p=8/9$ and computes two specific probabilities; (b) uses the given foci and latus-rectum length to find $a^2,b^2$ and write the ellipse equation. Both alternatives answered below.

(a) Binomial probabilities for ships arriving safely

1. Set up the binomial model. "1 ship out of 9 was sunk on average" means $P(\text{sunk})=q=\dfrac19$, so $P(\text{arrives safely})=p=1-\dfrac19=\dfrac89$. For $n$ independent ships, the number arriving safely $X\sim B(n,p)$ with $p=\dfrac89$.

2. (i) Exactly $3$ safe out of $6$, i.e. $n=6,\,X=3$.

$P(X=3)=\binom63p^3q^3=20\left(\dfrac89\right)^3\left(\dfrac19\right)^3$

3. $\left(\dfrac89\right)^3=\dfrac{512}{729}$, $\left(\dfrac19\right)^3=\dfrac1{729}$, so

$P(X=3)=20\times\dfrac{512}{729}\times\dfrac1{729}=\dfrac{10240}{531441}\approx0.01927$

4. (ii) No ships safe out of $4$, i.e. $n=4,\,X=0$ (all $4$ sunk):

$P(X=0)=\binom40p^0q^4=q^4=\left(\dfrac19\right)^4=\dfrac1{6561}\approx0.0001524$

(b) Ellipse with foci $(2,1),(-2,1)$ and latus rectum $6$

1. Centre. Midpoint of the foci: $\left(\dfrac{2+(-2)}2,\dfrac{1+1}2\right)=(0,1)$. The foci lie on the horizontal line $y=1$, so the major axis is horizontal.

2. $c$ value. Distance from centre to each focus: $c=2$.

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