Question

Suppose a family has 4 children, named a; b; c and d, who take turns washing 4 plates denoted p1; p2; p3; p4.
These children are not so careful in their work so, over time, each of the plates will be broken. Suppose any child
could break any plate and that all the arrangements in which plates p1; p2; p3; p4 could be broken by children
a; b; c; d are equally likely.

(a) Find the probability that child a breaks 3 plates.

(b) Find the probability that one of the four children breaks 3 plates

Answer 1

That is \[P _{1},P _{2},P _{3}, P _{4}\]

Answer 2

i can take a shot at this, you can tell me if it seems reasonable

Answer 3

all these words seem to indicate that the probability any child breaks any plate is \(\frac{1}{4}\) that is, the probability that Annie breaks plate 1 is \(\frac{1}{4}\) and the probability that she breaks plate 2 is \(\frac{1}{4}\) etc

Answer 4

we want to know what is the probability that she breaks 3 out of 4 plates.
the probability she breaks plate 1 and 2 and 3 and not plate 4 is \((\frac{1}{4})^3\times \frac{3}{4}\)

Answer 5

there are 4 different ways for her to break 3 plates, all with the same probability
so my guess at an answer is
\[4\times (\frac{1}{4})^3\times \frac{3}{4}=(\frac{1}{4})^2\times \frac{3}{4}=\frac{3}{64}\]

Answer 6

as for part 2, if part one is correct, the event that Anne breaks 3 plates is the same as the probability that Bethany, Chrystal and Dagwood do, and since these events are disjoint, you can add the probabilities
get \[\frac{12}{64}\]
if this seems unlikely let me know

Answer 7

I think you maybe correct I have the same logic just wanted a second opinion :)

Answer 8

glad to provide it
lot of verbiage in this question, isn't there?

Answer 9

oh yes and probability can be so confusing if worded the least bit out of the comfort zone