M41-70

Question

There are two fair dice, each numbered 1 to 6. If the dice are thrown  n  times, what is the probability of obtaining at least one double "4" in the  n  throws?

A.  1-(\frac{1}{36})^n 
B.  1-(\frac{25}{36})*n 
C.  1-(\frac{25}{36})^n 
D.  1-(\frac{35}{36})^n 
E.  (1-\frac{35}{36})^n

Bunuel · Answer

Official Solution:

There are two fair dice, each numbered 1 to 6. If the dice are thrown  n  times, what is the probability of obtaining at least one double "4" in the  n  throws?

A.  1-(\frac{1}{36})^n 
B.  1-(\frac{25}{36})*n 
C.  1-(\frac{25}{36})^n 
D.  1-(\frac{35}{36})^n 
E.  (1-\frac{35}{36})^n

The probability of getting a double "4" is  \frac{1}{6}*\frac{1}{6}=\frac{1}{36} .
 
The probability of NOT getting a double "4" is  1 - \frac{1}{36}=\frac{35}{36} .
 
The probability of NOT getting a double "4" in  n  throws is  (\frac{35}{36})^n .
 
Therefore, the probability of obtaining at least one double "4" in the  n  throws is  1 - (\frac{35}{36})^n .

Answer: D

The8 · Answer

Boenuel,

I have one question, initially I started approaching this question with what would be probability of not getting double for and calculating 1*5/6 (not getting 4 on the second dice) but did not arrive at the correct answer,

what is the flaw in my reasoning, can you help?

Bunuel · Answer

1 * 5/6 represents the probability of not having a 4 on the second die. However, with this approach, you exclude more cases than needed: (1, 4), (2, 4), (3, 4), (5, 4), and (6, 4). Adding the probability of these five cases gives 5/6 + 5/36 = 35/36. This result can be derived more straightforwardly: there are 36 total combinations when rolling two dice, and only one of them is (4, 4). Thus, the probability of not getting (4, 4) is 35/36.

The8 · Answer

All clear , thank you

OmerKor · Answer

Hi Bunuel

Thank you for your explanation of the dice problem.
I think I understand the complement method now, but I still feel some confusion about the definition of the event and how the trials are treated.

Specifically, here’s where I got stuck:
When n=2,
I imagined rolling 4 dice “all at once.”
Then I thought: if I get two 4’s anywhere among those four dice (for example, (4,2,4,6)), that should count as a “double-4.”
Based on that, I tried to calculate using combinations of positions for the 4’s, and I ended up with a probability of
1/(6^3).

But in your approach, the probability is 1−(35/36)^2 = 71/1296.
This excludes cases like (4,2,4,6)

I thought to ask you these 2 question to better understand the solution

1. Why do we define the event strictly as “both dice showing 4 within a single trial,” rather than “two 4’s appearing anywhere among the total 2n dice”?
2. Is the structure of the problem (phrasing: “the dice are thrown n times”) what forces us to treat each pair of dice as a separate event?

Thank you!

1. Why do we define the event strictly as “both dice showing 4 within a single trial,” rather than “two 4’s appearing anywhere among the total 2n dice”? 2. Is the structure of the problem (phrasing: “the dice are thrown n times”) what forces us to treat each pair of dice as a separate event? Thank you!

Bunuel · Answer

A “double-4” means both dice show 4  in the same throw of 2 dice . It does  NOT  mean “any two 4’s among all 2n dice.” Your (4,2,4,6) example is two throws, each with one 4, but never both at once, so no double-4.
 
 The phrasing “the dice are thrown n times” forces us to treat each pair of dice as one independent trial. Each trial has P(double-4) = 1/36, so overall it’s 1 − (35/36)^n.

M41-70

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