The minimum number of boxes a participant needs to eliminate is 12. The half chance of holding at least $100,000 as his/her chosen box is 7/14 from the whole chances of 7/26.
<h3>At what chance the participant can hold a box with $100,000?</h3>
There are 26 boxes.
The required money is at the 7 the stage box.
So, the chance of holding at least $100,000 is 7/26.
<h3>What is the minimum number of boxes a participant needs to eliminate for the required amount with half chance of holding?</h3>
Since the required value is available at a 7/26 chance of holding, the boxes that are less than $100,000 are removed.
So, 12 boxes holding less than $100,000. So, the probability of half chance of holding at least $100,000 is 7/14.
Learn more about how to calculate probability chances here:
brainly.com/question/25870256
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Disclaimer: The given question on the portal is incomplete. Here is the complete question.
Question: In the game Deal or No Deal, participants choose a box at random from a set of one containing each of the following values:
$0.01, $1,000
$1, $5,000
$5, $10,000
$10, $25,000
$25, $50,000
$50, $75,000
$75, $100,000
$100, $200,000
$200, $300,000
$300, $400,000
$400, $500,000
$500, $750,000
$750, $1,000,000
After choosing a box, participants eliminate other boxes by opening them, showing the amount of money in the box to the crowd, and then removing that box (and its money!) from the game. What is the minimum number of boxes a participant needs to eliminate to have a half chance of holding at least $100,000 as his or her chosen box?
