Other methods, such as the one suggested in @Noah's comment, may be preferable. However, to answer your
question directly, you could use a chi-squared test, provided
that you have enough data for the expected counts
to be large enough.
Suppose your counts were as follows for 100 Stroke patients and 1000 patients with no stroke.
Fruits/day 0 1 2 3 4+
Stroke 33 37 22 7 1
NonStr 181 347 305 130 37
Because the expected counts are mostly above 5 and
the remaining one is above 3, you could do a chi-squared
test to see whether the probabilities of various
amounts of fruit per day are the same for the two
groups of patients. (You'd still get a significant
result by combining categories
4+.) Output from R:
Pearson's Chi-squared test
X-squared = 17.307, df = 4, p-value = 0.001685
In chisq.test(DTA) : Chi-squared approximation
may be incorrect
The warning message is because of the one one expected
count below 5.
Notice, however, that this test does not take into
account that the Fruit variable is ordinal.
Two additional possibilities for such large numbers of subjects:
Using actual counts, both a Welch two-sample t test and a two-sample Wilcoxon test give essentially 0 P-values. Strictly speaking,
assumptions for neither test are met, but large sample sizes and tiny P-values show strong evidence of a
difference---for my fake data.
Note: Fake data were generated in R as:
x1 = rbinom(100, 5, .2)
x2 = rbinom(1000, 5, .3)
Roughly, the model is that people have five
opportunities a day to eat fruit, breakfast, lunch, dinner, and a couple of snacks. At any
one opportunity, respective probabilities of
doing so are $0.2$ and $0.3.$