Problem involving the binomial and normal distributions So we are given a cylinder with a fixed diameter of 62mm and length 616mm. There are also spherical balls of diameter 61.5mm and s.d. 0.2mm that are normally distributed. The question asks for the probability of successfully fitting 10 of these balls in the cylinder, by width and length.
I started by calculating the probability that the cylinder is long enough for 10 balls:
$P(L<616)=ncdf(-10^{99},616,61.5(10),0.2\sqrt{10})\approx 0.943$
Then I calculated the probability that the cylinder is wide enough for one ball:
$P(W<62)=ncdf(-10^{99},62,61.5,0.2)\approx 0.994$
I increased this answer to the power of 10, as there are 10 balls:
$[P(W<62)]^{10}=[0.944]^{10}\approx0.940$
To get the final answer I multipled the probabilities of width and length together:
$P($10 balls fit in cylinder$)\approx0.940\times0.943\approx0.886$
But I'm told my answer is wrong that and that I should not have increased the width probability to the power of 10, which gives the correct answer of:
$P($10 balls fit in cylinder$)\approx0.994\times0.943\approx0.937$
Can someone please explain why?
 A: Here is an approximate result via Monte Carlo simulation.  Code in R:
 itfits <- function() {
  balls <- rnorm(10, 61.5, 0.2)
  return(max(balls) <= 62 && sum(balls) <= 616)
}
nreps <- 1e6
set.seed(12344321)  # for reproducibility of results
x <- replicate(nreps, itfits())
print(sum(x))
prop.test(sum(x), nreps)

Out of $10^6$ trials, the balls fit in the tube in $895,997$ cases, giving an estimate of $0.895997$ of the probability of success, with a $95\%$ confidence interval of $0.8954$ to $0.8966$.
A: It is right to increase  the width probability to the power of 10, since every width of the ten balls must not exceed  62 mm.
And for the length we have a sum of 10 i.i.d. normally distributed random variables. Let $X_i$'s the random variables for the diameter of the balls. The 10 balls fit in the cylinder, if the sum of them is below  616. The sum of the 10 random variables $S_{10}$ is distributed as
$$S_{10}\sim\mathcal N\left(10\cdot 61.5, 10\cdot 0.2^2  \right)$$
Then in case of the length of the cylinder it asked for $P(S_{10}\leq 616)$. A picture can help to understand better the situation.

