Calculate conditional probabilities Statistics & Probabilities

The probability of getting a blue marble on the first pick is:

P\left(A\right)=\dfrac{7}{12}

After the first pick, there are only 6 blue marbles left and 11 marbles in total. Therefore, the probability of getting a blue marble on the second pick, given that the first picked marble was blue, is:

P\left(B/A\right)=\dfrac{6}{11}

Therefore, the probability of getting two blue marbles is:

P\left(A\right)\cdot P\left(B/A\right)=\dfrac{7}{12}\cdot\dfrac{6}{11}=\dfrac{7}{22}

The probability of getting a green ticket on the first pick is:

P\left(A\right)=\dfrac{10}{16}=\dfrac{5}{8}

After the first pick, there are only 9 green tickets left and 15 tickets in total. Therefore, the probability of getting a yellow on the second pick, given that the first picked ticket was green, is:

P\left(B/A\right)=\dfrac{6}{15}=\dfrac{2}{5}

So, the probability of getting a green ticket on the first pick and a yellow ticket on the second pick is:

P\left(A\right)\cdot P\left(B/A\right)=\dfrac{5}{8}\cdot\dfrac{2}{5}=\dfrac{1}{4}.

The probability of getting a green marble on the first pick is:

\dfrac{10}{26}=\dfrac{5}{13}

After the first pick, there are only 9 green marbles left and 25 marbles in total. Therefore, the probability of getting a green marble on the second pick, given that the first picked marble was green, is:

\dfrac{9}{25}

After the second pick, there are only 8 green marbles left and 24 marbles in total. Therefore, the probability of getting a green marble on the third pick, given that the first picked marble was green and the second picked marble was green, is:

\dfrac{8}{24}=\dfrac{1}{3}

In conclusion, the probability of George getting three green marbles is:

\dfrac{5}{13}\cdot\dfrac{9}{25}\cdot\dfrac{1}{3}=\dfrac{3}{65}

Suppose that:

• A is the event of drawing an ace in the first pick.
• B is the event of drawing an ace in the second pick.

There are 4 aces in the 52 cards:

\dfrac{4}{52}=\dfrac{1}{13}

If we pick an ace on our first try, then there are only 3 aces left and 51 cards in total. The probability of drawing an ace on the second pick, given that the first card was an ace is:

P\left(B|A\right)=\dfrac{3}{51} = \dfrac{1}{17}

The probability of drawing two aces equals:

P\left(A \cap B\right) = P\left(A\right).P\left(B|A\right) = \dfrac{1}{13}\times\dfrac{1}{17} =\dfrac{1}{221}

Suppose that:

• A is the event of getting a six, which is one of the following cases:

(1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (6,1), (6,2), (6,3), (6,4), (6,5)

• B is the event of getting two different numbers, which is the set of all possible pairs excluding the pairs with the same numbers:

(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

Hence:

P\left(B\right) = \dfrac{30}{36}=\dfrac{5}{6}

• Excluding (6,6) from the event A, we have the event A \cap B. Therefore:

P\left(A \cap B\right)=\dfrac{10}{36}= \dfrac{5}{18}.

Therefore, the probability of getting a six given that the numbers are different equals:

P\left(A | B\right) =\dfrac{P\left(A \cap B\right) }{P\left(B\right)}=\dfrac{5}{18} \div\dfrac{5}{6}=\dfrac{1}{3}

Suppose that:

• A is the event that at least one of the children is a girl \left(gg,gb,bg\right), so: P\left(A\right)=\dfrac{3}{4}
• B is the event that both children are girls (gg), so:

P\left(A\cap B\right)=\dfrac{1}{4}

We know that:

P\left(B|A\right) = \dfrac{P\left(A \cap B\right)}{P\left(A\right)}=\dfrac{1}{4}\div \dfrac{3}{4}=\dfrac{1}{3}

Suppose that:

• A is the event of getting a prime number, namely 2,3,5
• B is the event of getting 5.

Our aim is to compute P\left(B|A\right)

We know that:

P\left(A\right)=\dfrac{3}{6}=\dfrac{1}{2}

and:

P\left(A \cap B\right) = \dfrac{1}{6}

We have:

P\left(B|A\right) = \dfrac{P\left(A \cap B\right)}{P\left(A\right)}=\dfrac{1}{6} \div\dfrac{1}{2}=\dfrac{1}{3}