Answer: 0.33
Explanation:
Let C be the event of ordering chicken sandwich.
F be the event of ordering french fries.
As per given , we have
P(C)=0.65
P(F)=0.44
P(C∩F)=0.76
Formula we use here,

Then, For given situation, we have

i.e.

Hence, the probability that a randomly selected customer will order both a chicken sandwich and french fries=0.33