Jesse has eight friends who have volunteered to help with a school fundraiser. Five are boys and three are girls. If he randomly selects 3 friends to help him, find each probability? Probability of all boys or all girls
The probability of choosing a boy on the first selection is 5/8. Having chosen a boy on the first selection, there are 4 boys out of a total of 7 friends remaining. Therefore the probability of choosing a boy on the second selection is 4/7. Having chosen boys on the first and second selections, there are 3 boys out of a total of 6 friends remaining. Therefore the probability of choosing a boy on the third selection is 3/6. \[\large P(all\ boys)=\frac{5\times4\times3}{8\times7\times6}=you\ can\ calculate\]
Okay since the question is asking for all boys or all girls Do I do the same for the girls and add them together
The events 'all boys' and 'all girls' are mutually exclusive, therefore you need to calculate the probability of all girls (using the same reasoning as for all boys) and add the two values of probability.
I got it It is 11/56 I have another question involving this problem it is asking for probability of at least 2 girls
You need to find the probability of exactly 2 girls. Then add that probability to the probability of all girls.
So 2 out of 8 + 3/8 Is 5/8
Not so. The probability of all girls is given by \[\large P(all\ girls)=\frac{3\times2\times1}{8\times7\times6}=you\ can\ calculate\] The probability of exactly 2 girls is given by \[\large P(2\ girls)=\frac{3C2\times5C1}{8C3}=\frac{3\times5}{56}=you\ can\ calculate\]
Okay so 1/56 plus 15/56 Is 16/56 simified is 4/14
And 4/14 simplified is 2/7.
I have one more Probability of at least 1 boy It's exact of 1 boy plus Al Boys correct So 1/56 plus what?
The probability of no boys is 1/56. Therefore the probability of at least 1 boy is given by \[\large P(at\ least\ 1\ boy)=1-1/56=you\ can\ calculate\]
So 0/56? But my book says it 55/56
How did you get 0/56? 1 - 1/56 = 55/56
Oh my bad I didn't do it correct Thank for helping me I appreciate it
You're welcome :)
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