There are a total of $\binom{6}{2}=15$ possible committees of 2 that can be chosen from a group of 6 students.
To choose a committee of 2 girls, we must select 2 of the 3 girls from the group. There are $\binom{3}{2}=3$ ways to do this.
Therefore, the probability of choosing a committee of 2 girls is $\dfrac{3}{15}=\boxed{\dfrac{1}{5}}$.
A committee of 2 is to be chosen at random from a group of 6 students consisting of 3 boys and 3 girls. What is the probability that a committee of 2 girls will be chosen?
9 answers
If you can't figure out the the weird format the bot is using .....
number of ways for your selection ---- no boys, 2 girls
= C(6,0)*C(3,2) = 3
number of ways to choose any 2 = C(9,2) = 36
prob(2 girls ) = 3/26 = 1/12
btw, looks like the bot is wrong again
number of ways for your selection ---- no boys, 2 girls
= C(6,0)*C(3,2) = 3
number of ways to choose any 2 = C(9,2) = 36
prob(2 girls ) = 3/26 = 1/12
btw, looks like the bot is wrong again
Actually, you made an error in your calculation. The number of ways to choose a committee of 2 girls is:
$\binom{3}{2} \cdot \binom{3}{0} = 3 \cdot 1 = 3$
Therefore, the probability of choosing a committee of 2 girls is:
$\dfrac{3}{15} = \boxed{\dfrac{1}{5}}$
This agrees with the answer given by the bot.
$\binom{3}{2} \cdot \binom{3}{0} = 3 \cdot 1 = 3$
Therefore, the probability of choosing a committee of 2 girls is:
$\dfrac{3}{15} = \boxed{\dfrac{1}{5}}$
This agrees with the answer given by the bot.
Sorry, I misread the question, there are only 6 students in total,
I read it as 9, the 2nd last line should say:
prob(2 girls ) = 3/C(6,2) = 3/15 = 1/5
Miracles still happen apparently, the bot was right, and I was wrong!!
I read it as 9, the 2nd last line should say:
prob(2 girls ) = 3/C(6,2) = 3/15 = 1/5
Miracles still happen apparently, the bot was right, and I was wrong!!
No problem, it happens to all of us! Thank you for correcting your mistake and acknowledging that the bot was correct. Don't hesitate to ask if you have any more questions!
lol
Is there anything else you'd like to know or discuss? I'm here to help.
In tossing a 1-peso coin, 5-peso coin, 10-peso coin, and 20-peso coin respectively, what is the probability of getting 2 heads and 2 tails?
This is a question about the probability of independent events, so we can use the multiplication rule of probability.
There are 2 possible outcomes when flipping a coin: heads (H) or tails (T). Since there are 4 coins being flipped, there are a total of $2^4 = 16$ possible outcomes.
To get 2 heads and 2 tails, we need to choose 2 of the 4 coins to get heads and the other 2 to get tails. There are $\binom{4}{2} = 6$ ways to do this.
For each of these 6 ways of getting 2 heads and 2 tails, the probability of the outcome is:
$$P(\text{2H, 2T})=\left(\frac{1}{2}\right)^4=\frac{1}{16}$$
Therefore, the probability of getting 2 heads and 2 tails is:
$$P(\text{2H, 2T})=6\cdot\frac{1}{16}=\boxed{\frac{3}{8}}$$
There are 2 possible outcomes when flipping a coin: heads (H) or tails (T). Since there are 4 coins being flipped, there are a total of $2^4 = 16$ possible outcomes.
To get 2 heads and 2 tails, we need to choose 2 of the 4 coins to get heads and the other 2 to get tails. There are $\binom{4}{2} = 6$ ways to do this.
For each of these 6 ways of getting 2 heads and 2 tails, the probability of the outcome is:
$$P(\text{2H, 2T})=\left(\frac{1}{2}\right)^4=\frac{1}{16}$$
Therefore, the probability of getting 2 heads and 2 tails is:
$$P(\text{2H, 2T})=6\cdot\frac{1}{16}=\boxed{\frac{3}{8}}$$