Question

A basket of fruit contains 4 bananas, 3 apples, and 5 oranges. You intend to
draw a piece of fruit from the basket, keep it, draw a 2nd piece of fruit from the
basket, keep that, and then draw a third piece of fruit from the same basket. What
is the probability of selecting one banana, then one apple, then one orange, in
that order, from the basket if you are blindfolded?

Answer 1

Hi

draw 1 :  4/12 = 1/3

draw2 :  3/11

draw 3 : 5/10 = 1/2

so probability to have  :  1 banana then 1 apple then 1 orange  is :  1/3*3/11*1/2 =  3/66 =  3*1 / 2*3*11 =  1/22

Answer 2

Answer:

1/22

Step-by-step explanation:

Work out the probability of selecting each fruit then times them together so

there are 12 fruits in total and 4 bananas so the chance of selecting a banana is 4/12 simplified 1/3

then as you dont put the banana back there is only 11 fruits left in the basket and 3 are apples so the chance of selecting an apple is 3/11

then as you dont put the apple back there is only 10 fruit left in the basket and 5 are oranges so the chance of selecting an orange is 5/10 simplified 1/2

then we multiply them all together 1/3 x 3/11 x 1/2 3/66 to get which simplifies to 1/22