Probability

Let $X$ be the number of hits in 3 attempts.

Given: $p = P(\text{hit}) = 0.25,\qquad q = P(\text{miss}) = 1 - 0.25 = 0.75,\qquad n = 3$

Using binomial distribution, $P(X=x) = {}^nC_x p^x q^{n-x}$

a) Probability that none will hit the target
$P(X=0) = {}^3C_0 (0.25)^0 (0.75)^3$ $= 1 \times 1 \times 0.421875$ $= 0.421875$

b) Probability that exactly two will hit the target
$P(X=2) = {}^3C_2 (0.25)^2 (0.75)^1$ $= 3 \times 0.0625 \times 0.75$ $= 0.140625$

Answers: a) $0.421875$ b) $0.140625$

Let $A,B,C$ be the events that a cap is made by machine A, B, C, and $D$ the event that it is defective.

$P(A)=0.30,\ P(B)=0.35,\ P(C)=0.45$ $P(D|A)=0.02,\ P(D|B)=0.03,\ P(D|C)=0.04$

Total probability of a defective cap

$P(D)=0.30(0.02)+0.35(0.03)+0.45(0.04)=0.006+0.0105+0.018=0.0345$

Bayes' theorem

$P(A|D)=\frac{P(A)\,P(D|A)}{P(D)}=\frac{0.006}{0.0345}=\frac{4}{23}\approx0.174$

The probability that the defective cap came from machine A is $\dfrac{4}{23}\approx 0.174$.

(The given shares add up to 110%; the ratio above is unaffected by this, since each share is used only as a weight.)

Given: A fair coin is tossed $n = 4$ times.

To find: Probability of getting at least 2 heads.

Solution:

Total number of outcomes: $n(S) = 2^4 = 16$

Let $X$ be the number of heads. Using the binomial distribution with $p = \frac{1}{2}$ and $q = \frac{1}{2}$:

$P(X = r) = \binom{4}{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{4-r} = \frac{\binom{4}{r}}{16}$

Probability of at least 2 heads: $P(X \geq 2) = P(2) + P(3) + P(4)$

$= \frac{\binom{4}{2}}{16} + \frac{\binom{4}{3}}{16} + \frac{\binom{4}{4}}{16}$

$= \frac{6}{16} + \frac{4}{16} + \frac{1}{16} = \frac{11}{16}$

Alternative method: $P(X \geq 2) = 1 - [P(0) + P(1)] = 1 - \left(\frac{1}{16} + \frac{4}{16}\right) = 1 - \frac{5}{16} = \frac{11}{16}$

Result: The probability of getting at least 2 heads is $\boxed{\dfrac{11}{16}}$.

(b) $P(A/B) = \dfrac{P(A \cap B)}{P(B)}$

The conditional probability of A given B is the probability of the intersection divided by the probability of the conditioning event B, so $P(A/B)=\dfrac{P(A\cap B)}{P(B)}$ (provided $P(B)\neq 0$). Option (a) would give $P(B/A)$, not $P(A/B)$.

For a binomial distribution with $n=5$: $P(X=r)={}^5C_r\,p^r q^{5-r}$ and $q=1-p$.

Finding p and q

$P(3)={}^5C_3\,p^3q^2=10p^3q^2=\frac{144}{625}$

Try $p=\frac{2}{5},\ q=\frac{3}{5}$:

$10\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^2=10\times\frac{8}{125}\times\frac{9}{25}=\frac{720}{3125}=\frac{144}{625}\ \checkmark$

Check with $P(1)=5pq^4=5\times\frac{2}{5}\times\frac{81}{625}=\frac{162}{625}$, which fits the same $p$ (the printed value $\frac14$ appears to be a misprint of $\frac{162}{625}$).

So $p=\frac{2}{5}$ and $q=\frac{3}{5}$.

Probability of 2 successes

$P(2)={}^5C_2\,p^2q^3=10\times\frac{4}{25}\times\frac{27}{125}=\frac{1080}{3125}=\frac{216}{625}\approx 0.3456$

Let $J$: visitor went to Janakpur, $L$: visitor went to Lumbini, $N$: visitor is Nepali.

Given $P(J)=0.70,\ P(L)=0.30,\ P(N\mid J)=0.35,\ P(N\mid L)=0.65$.

By Bayes' theorem:

$P(J\mid N)=\frac{P(J)P(N\mid J)}{P(J)P(N\mid J)+P(L)P(N\mid L)}=\frac{0.70\times0.35}{0.70\times0.35+0.30\times0.65}$

$=\frac{0.245}{0.245+0.195}=\frac{0.245}{0.440}=\frac{49}{88}\approx 0.557$

So the probability that the Nepali visitor went to Janakpur is $\frac{49}{88}\approx0.557$.