It is $P(X\geq 5)=P(X=5)+P(X=6)$, where $X\sim Bin \left(6, \frac46 \right)$ Comprehensible? If yes, then it´s your turn.

Nope callculus 🤔

What is the probability that a die (singular) show at least 5 if it is thrown once?

I think that will be: 5/6

@Ador There are 6 possible outcomes: $1,2,3,4,5,6$, The favorable outcomes are: 5 and 6. So the probability is ...

Not really. The denominator of 6 is right. But the numerator (favorable outcomes) is what?

It should be: 2/6 = 1/3

@Ador That´s true. And the probability that at least 5 rolls show at least a 5 is $$P(X\geq 5)=\sum_{k=5}^6\binom{6}{k}\cdot \left( \frac13\right)^k\cdot \left( 1-\frac13\right)^{6-k}=\frac{13}{729}\approx 1.78\%$$

Thanks a bunch "callculus" bro 💖 Would you mind giving me the link of your social media account?

@Ador You´re welcome. Usually I don´t post my accounts in order to protect my privacy-sorry.

@drhab A side note to the closer: Ador is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. My question: Why did you vote to close?

Callculus bro do you know any other math help forum except this?

?

Not really. I have no experience with other forums (in english). So I cannot recommend another one.

You entered a room of 34 people. A shooter then enters killing 30. What is the probability of people that are left in the room?

Callculis, bro what will be the answer of this problem?