-4
$\begingroup$

Please help me solve this

The probability of $3$ dart players Evelyn, Tracy and John hitting the bulls eye are $0.2$, $0.3$ and $0.5$ respectively.

(I) Find the probability that all of them hit the bull's eye
(II) Only one of them hits the bull's eye
(III) At most one of them missed the bull's eye

$\endgroup$
2
  • 4
    $\begingroup$ You should really put in your attempt and what you dont get about the problem in the question. $\endgroup$
    – ʎpoqou
    May 20, 2018 at 9:29
  • $\begingroup$ Assuming their trials are independent. (I)$P(\text{everyone hits a bulls eyes})$=0.2x0.3x0.5. (II) 0.2x0.7x0.5+0.3x0.8x0.5+0.5x0.8x07 (III) Can you try figure it out? Add probabilities that all of them hit it, that Tracy and John hit it but not Evelyn and 2 more cases. $\endgroup$
    – ʎpoqou
    May 20, 2018 at 9:32

2 Answers 2

0
$\begingroup$

I) Assuming that they are independent events you just apply the independence rule using the given probabilities.

P(E & T & J) = P(E) * P(T) * P(J) = 0.2 * 0.3 * 0.5 = 0.03

II) What you have to do is figure out the different combinations of hits and misses that results in only one of them hitting the bull's eye.

So we have: 1) Ev hits, Tracy misses and John misses. 2) Ev misses, Tracy hits and John misses. 3) Ev misses, Tracy misses and John hits.

Then you calculate the probability of each of those combinations happening using the probabilities you've been given. and you add them together. So for 1) It's 0.2 * 0.7 * 0.5 = 0.07. 2) It's 0.8 * 0.3 * 0.5 = 0.12. 3) It's 0.8 * 0.7 * 0.5 = 0.28. Then, finally, 0.07 + 0.12 + 0.28 = 0.47.

III) I'm pretty sure that you do the same thing you did in (II) but you find the combinations that involve one person missing and the combinations that involve no one missing. Then you add them together.

(II) and (III) are easy if you draw a probability tree and follow the branches along to find the probability of each combination and add them at the end.

$\endgroup$
0
$\begingroup$

An important, but unstated, assumption is that each person's chance of hitting the bulls eye is independent of the others'.

Here is an approach that uses probability generating functions. Wikipedia on probability generating functions

Evelyn's probability generating function is $E(x) = 0.8 + 0.2 x$, Tracy's function is $T(x) = 0.7 + 0.3 x$, and John's function is $J(x) = 0.5 + 0.5 x$. It should be clear how these functions are derived from the problem statement. Here's the neat thing: if we multiply the three generating functions together and expand the product, the coefficient of $x^n$ in the resulting polynomial is the probability of getting a total of exactly $n$ hits on the bulls eye. If you think about how multiplication of polynomials works, I think you will see why. So consider $$ E(x) \cdot T(x) \cdot J(x) = 0.28\, +0.47 x+0.22 x^2+0.03 x^3$$

This function answers all the questions you might have about the total number of hits. The probability of zero hits is the coefficient of $x^0$, $0.28$; the probability of exactly one hit is the coefficient of $x^1$, $0.47$; and so on.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.