Exercise 3: Mobile Users and Checkout Completion — Possible Solution
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GIVEN
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P(mobile) = 0.6
P(checkout) = 0.1
P(mobile and checkout) = 0.03

STEP 1: P(checkout | mobile)
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Per this chapter's own conditional probability formula:

  P(checkout | mobile) = P(mobile and checkout) / P(mobile)
                        = 0.03 / 0.6
                        = 0.05 (5%)

STEP 2: COMPARING TO THE OVERALL P(checkout)
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P(checkout | mobile) = 5%
P(checkout) overall   = 10%

Since 5% is LESS than the overall 10% checkout rate, mobile users are
LESS likely than average to complete checkout - conditioning on
"mobile" genuinely lowers the probability compared to the unconditioned
baseline.

STEP 3: THE INDEPENDENCE TEST
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P(mobile) x P(checkout) = 0.6 x 0.1 = 0.06
P(mobile and checkout) = 0.03

Since 0.06 does not equal 0.03, per this chapter's own multiplication-
rule test, "on mobile" and "completes checkout" are NOT independent -
they are dependent events. Being on mobile measurably changes the
probability of completing checkout (and specifically lowers it),
which is exactly what Step 2's direct comparison already showed from
a different angle.

WHY THIS WORKS AS AN ANSWER
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The conditional probability is computed directly from this chapter's
own formula, compared explicitly against the unconditioned baseline to
draw a concrete practical conclusion (mobile users convert less), and
the independence test is applied as a second, independent confirmation
that these two events are genuinely related rather than coincidentally
appearing related.