Question 1155267
<br>
P(first chicken burger is the 4th sold) = P(1st AND 2nd AND 3rd are NOT chicken AND 4th is)<br>
{{{((97/155)^3)(58/155)}}}<br>
P(first chicken burger is the 5th sold) = P(1st AND 2nd AND 3rd AND 4th are NOT chicken AND 5th is)<br>
{{{((97/155)^4)(58/155)}}}<br>
You can do the calculations and express the answer in any form you want.<br>
Note my interpretation of the given numbers is that 97 of every 155 burgers sold are beef or vegetable and 58 of every 155 are chicken, so the probability of any given burger being chicken is 58/155 and the probability of its being beef or vegetable is 97/155.<br>
Changing the probabilities as burgers are sold seems unreasonable, since it would suggest that there are EXACTLY 58 chicken burgers and EXACTLY 97 beef or vegetable burgers sold each day.<br>