Getting into college Ramon has applied to both Princeton and Stanford. According to his counselor, the probability that Princeton will admit him is $0.4$, the probability that Stanford will admit him is $0.5$, and the probability that both will admit him is $0.2$

(a) Make a Venn diagram to model this chance process.

(b) What is the probability that neither university admits Ramon?

(c) What’s the probability that he gets into at least one of the two schools? Use the general addition rule to confirm that your answer is correct.

• When appropriate, use a tree diagram to describe chance behavior.
• Use the general multiplication rule to solve probability questions.
• Compute conditional probabilities.

Princeton has a $0.4$ percent chance of admitting him. Stanford has a $0.5$ percent chance of admitting him. And there's a $0.2$ chance that both of them will admit him.

A tree diagram can be used to depict the sample space when chance behavior involves a series of outcomes. Tree diagrams can also be used to determine the likelihood of two or more events occurring at the same time. Simply multiplying along the branches that correspond to the desired results is all that is required.

Draw two ellipses that partially overlap. The one on the left is "Princeton," and the one on the right is "Stanford." In the intersection of the ellipses, write the likelihood of being admitted to both. Being admitted to Princeton has a probability of $0.4$, whereas being admitted has a probability of $0.2$, hence the likelihood of being admitted to Princeton alone is $0.2$ Because the chance of being admitted to Stanford is $0.5$ and the chance of being admitted is $0.2$, the chance of being admitted to Stanford alone is $0.2$ To the right of the intersection, this must be written.

Outside of the two ellipses, the likelihood of not being admitted to either university is expressed as $0.3$

Addition rule: $P(AorB)=P\left(A\right)+P\left(A\right)-P(AandB)$

Using addition rule: $P(AorB)=P\left(A\right)+P\left(A\right)-P(AandB)$

As a result, the following was obtained:

$$\begin{gathered} P(PrinctonorS\tan dford)=P\left(Princton\right)+P\left(S\tan dford\right)-P(WhiteandHispanic)-P(PrinctonandS\tan dford) \\ =0.4+0.5-0.2 \\ =0.7 \end{gathered}$$