Answer:
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:
- Case 1, Where Hannah gets prize when she buys the first box:
Let K be the event of Hannah winning the prize on buying the first box.
⇒P(K)=P(A)=0.4
- Case 2, Where Hannah gets prize when she buys the second box:
I<u>n this event Hannah should not get the prize in first box but should get the prize on buying the second box</u>
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
- Case 3,Where Hannah gets prize when she buys the third box:
<u>In this event Hannah should not get the prize in first and second box but should get the prize on buying the third box</u>
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
