# Thread: Basic Probability

1. ## Basic Probability

These are three questions on the subject of basic probability. I have answers, could you please check them?

A bag contains 12 sweets of which 5 are red, 4 are green and 3 are yellow.
A boy chooses 3 of these sweets at random without replacement. Find the probability that
(a) he chooses 3 red sweets, [2]
(b) he chooses no red sweets, [2]
(c) he chooses 1 sweet of each colour. [3]

An A-level Mathematics class contains 5 boys and 6 girls. The teacher is told that 3 members of the class can go to a special lecture and she decides to select these 3 members at random. Find the probability that
(a) 2 boys and 1 girl are selected, [3]
(b) the 3 members selected are all of the same gender. [4]

Two unbiased cubical dice are thrown simultaneously. Calculate the probability that
(a) the score on each die is at least 3, [3]
(b) the scores on the two dice differ by 3. [3]

1]
(a) 5/12 x 4/11 x 3/10 = 60/1320 --> 1/22
(b) 7/12 x 6/11 x 5/10 = 210/1320 --> 7/44
(c) 5/12 x 4/11 x 3/10 = 1/22 [there are six different ways of doing this, so 1/22 x 6 = 6/22, or 3/11]

2]
(a) 5/11 x 4/10 x 6/9 = 12/99 --> 4/33 [again, there are six different ways so 6 x 4/33 = 24/33]
(b) 5/11 x 4/10 x 3/9 = 6/99 --> 2/33
and 6/11 x 5/10 x 4/9 = 4/33
so 2/33 + 4/33 = 6/33

3]
(a) 4/6 x 4/6 = 16/36 --> 4/9
(b) 6/36 --> 1/6

Thank you if you can help it is greatly appreciated.

2. Seems like all answers are correct.
good job

3. Hello, db5vry!

Sorry, I don't agree with one of your answers.
. . But the rest are correct . . . Nice work!

2) An A-level Mathematics class contains 5 boys and 6 girls.
The teacher selects 3 members of the class can go to a special lecture.

Find the probability that:
. . (a) 2 boys and 1 girl are selected.

My work:

$(a)\;\frac{5}{11}\times\frac{4}{10}\times\frac{6}{ 9} \:=\:\frac{4}{33}$

$\text{Again, there are six different ways . . . }\quad{\color{blue}\text{ no}}$

There are only three different ways: . $BBG,\:BGB,\:GBB$

The probability is: . $\frac{4}{33} \times 3 \:=\:\boxed{\frac{4}{11}}$

Let's solve it a different way . . .

There are 11 students, 3 are chosen.
. . There are: . $_{11}C_3 \:=\:\frac{11!}{3!\,8!} \:=\:{\color{blue}165}$ possible outcomes.

We want 2 of the 5 boys: . $_5C_2 \:=\:10$ ways.
We want 1 of the 6 girls: . $_6C_1 \:=\:6$ ways.
. . Hence, there are: . $10\cdot6 \:=\:{\color{blue}60}$ ways to choose 2 boys and a girl.

Therefore: . $P(\text{2 boys, 1 girl}) \:=\:\frac{60}{165} \:=\:\boxed{\frac{4}{11}}$