Plz me! whoever gets it right will be marked brainiest

Probability that Hannah only has to buy 3 or less boxes before getting a prize is 0.784

Step-by-step explanation:

Given, 40% of cereal boxes contain a prize

⇒probability of getting a prize on opening a box, P(A)=0.4

where A is the event of getting a prize on opening a cereal box

and probability of not getting a prize on opening a box, P(A')=1-P(A)=0.6

where A' is the event of not getting a prize on opening a cereal box

This problem needs to be divided into 3 situation:

Let K be the event of Hannah winning the prize on buying the first box.

⇒P(K)=P(A)=0.4

In this event Hannah should not get the prize in first box but should get the prize on buying the second box

Let L be the event of Hannah winning the prize on buying the second box

So, P(L)=P(A')·P(A)

           =(0.6)·(0.4)

           =0.24

In this event Hannah should not get the prize in first and second box but should get the prize on buying the third box

Let L be the event of Hannah winning the prize on buying the third box

So, P(L)=P(A')·P(A')·P(A)

           =(0.6)·(0.6)·(0.4)

           =0.144

Let N be the event of Hannah winning the prize on buying 3 or less boxes before getting a prize

⇒N=K∪L∪M

Now, Required probability is P(N)=P(K∪L∪M)=P(K)+P(L)+P(M) [As events K,L and M are independent and disjoint events]

⇒P(N)=0.4+0.24+0.144

         =0.784

ou would have 12 of each

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week