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

the correct answer to your question is a. 4^4

good luck my friend!

hope this !

maybe if i you could me by giving me brainliest?

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answer;

the correct answer

just use google you beautiful person.

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