Understanding Probability with Tree Diagrams
This video explains how to calculate probabilities and conditional probabilities using tree diagrams through several practical examples.
Example 1: Probability of Events A and B
- Given probabilities for events A and B and their complements (A', B') are shown.
- The sum of probabilities for an event and its complement equals 1.
- Probabilities along branches are multiplied to find joint probabilities, e.g., P(A ∩ B) = (1/5) * (1/4) = 1/20.
- Labels such as A ∩ B, A ∩ B', A' ∩ B, and A' ∩ B' are used to identify intersections.
- To find P(B), add probabilities of all branches containing B.
- Conditional probability P(A'|B) is calculated using the formula: [ P(A'|B) = \frac{P(A' \cap B)}{P(B)} ] Example: P(A'|B) = (12/40) ÷ (7/20) = 6/7.
Example 2: Jose's Bus and School Arrival Problem
- Probability Jose misses the bus (E) is 1/3; if missed, probability of being late (F) is 7/8.
- If he doesn't miss the bus (E'), probability of being late is 3/8.
- Tree diagram branches are multiplied to find joint probabilities, e.g., P(E ∩ F) = (1/3) * (7/8) = 7/24.
- To find P(F), sum probabilities of all branches containing F.
- Probability Jose misses the bus and is not late: P(E ∩ F') = 1/24.
- Conditional probability Jose missed the bus given he is late: [ P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{7/24}{13/24} = \frac{7}{13} ]
Example 3: Selecting Balls Without Replacement
- Bag A contains 3 white and 4 red balls; two balls are chosen without replacement.
- Probabilities adjust after each draw (e.g., second draw denominator decreases by 1).
- Probabilities for drawing two white balls: [ P(WW) = \frac{3}{7} * \frac{2}{6} = \frac{6}{42} ]
- Bag B contains 4 white and 3 red balls; probability of two white balls: [ P(WW) = \frac{4}{7} * \frac{3}{6} = \frac{12}{42} = \frac{2}{7} ]
Example 4: Combined Probability with Dice Roll
- A standard die is rolled:
- If 1 or 2 is rolled, two balls are chosen from Bag A.
- Otherwise, two balls are chosen from Bag B.
- Probability of rolling 1 or 2 is 2/6; rolling other numbers is 4/6.
- Probability of two white balls overall: [ P(WW) = P(roll 1 or 2) * P(WW|A) + P(roll other) * P(WW|B) ] [ = \frac{2}{6} * \frac{6}{42} + \frac{4}{6} * \frac{2}{7} = \frac{1}{21} + \frac{4}{21} = \frac{5}{21} ]
- Conditional probability that balls came from Bag B given both are white: [ P(B|WW) = \frac{P(B) * P(WW|B)}{P(WW)} = \frac{\frac{4}{6} * \frac{2}{7}}{\frac{5}{21}} = \frac{1}{5} ]
Key Takeaways
- Tree diagrams help visualize and calculate joint and conditional probabilities.
- Multiplying probabilities along branches gives joint probabilities.
- Conditional probabilities use the formula P(A|B) = P(A ∩ B) / P(B).
- Adjust probabilities when sampling without replacement.
- Combining multiple stages (like dice rolls and ball draws) requires multiplying probabilities across stages and summing over possible paths.
This approach is essential for solving complex probability problems involving multiple dependent events.
For a deeper understanding of probability concepts, check out our resources on Introduction to Probability and Statistics: Key Concepts and Terminology and Understanding Z-Scores and their Applications in Statistics. Additionally, you can explore Comprehensive Review of Discrete Probability Distributions and Expected Values for more examples and applications.
the diagram below shows the probabilities for Invent A and B with the p of a prime equals P write down the
value of p p = 1 - 1 over 5 = 4 over 5 because one over five and a p come from the same note the sum of this two
probabilities equals one then we will figure out the probabilities by
multiplying the probabilities along the branches 1/ 5 * 1/ 4 = 1 1/
20 1/ 5 * 3 4 = 3 over 20 4 over 5 4 over 5 * 3 over 8 equal 12 over 40 you can use calculator to reduce or you can
leave as it is 4 over 5 * 5 8 = 20 over
40 then we need to label a b a b
prime a prime b a prime or B Prime basically this means the a intersection
B this means the a in intersection B Prime this means a prime or intersection B this means a prime or intersection B
Prime B find the P of B we will check the labels as long as it contains the B will
be the solutions contains B contains B so we need to add this two
probabilities together probabilities of b equals 1/ 20 + 12 over
40 = 7 over 20 say find the
P of a prime given B we will use this formula to figure out P of a prime or given
b equals P of a prime intersection B all over probability of b equals a prime or intersection B is
this 12 over 40 probability of B is here 7 over 20 12 over
40 over 7 over 20 = 6 over 7 Jose travels to school on a bus
on any day the probability that Jose will miss the bus is 1 over three if he misses his bus the probability that he
will be late for school is 7 over 8 if he doesn't miss his bus the probability that he will be late it's a
3 over 8 let ye be B the invent he misses his bus and F the invent he is late for
school the information above is shown on the tree diagram this e
means Mr the bus f means late for
school we will figure out the probabilities by multiplying
probabilities along the branches 1 over 3 * 7/ 8 = 7/ 24 1/ 3 * 1/ 8 = 1/
24 2 over 3 * 3 over 8 = 6 over 24 2 over 3 * 5/ 8 = 10 over 24 then we need to
label e f e f Prime e Prime
f e prime or F prime a find the P of e intercept section f e intersection f 7/
24 probability of f we will check the labels containing F then we need to
add these two probabilities together to get this probability of for f which is 7 over 24
+ 6 over 24 equal 13 over 24 B find the probability that Jose misses his b and it's not late for
school misses his pass means e and means intersection not late for school is f
Prime so we are looking for probability of for E intersection F
Prime let's check the label e intersection F Prime is here 1 over 24 is the
solution B2 Jose missed his bus given given that he is late for school given that this is a conditional
probability late for school late for school means F probability of f will go to
denominator let's check back the probability of f it's here we need to check these two
probabilities to decide which one go to numerator given that this one will go to denominator for sure which means 13 over
24 go to denominator then tck here miss his pass 7 over 24
means Miss is a bass so 7 over 24 go to
numerator Final Answer equals 7 over 13 bag a contains three white balls and four red balls two balls are chosen at
Red without replacement copy and complete the following tree
diagram without replacement for the red then check this red because you will
make a numerator and a denominator both minus one for without replacement 3 over
6 then for this probability we will do one - 3/ 6 = 3/ 6 2 over
6 1 - 2/ 6 = 4 over 6 4 over 6 for second one find the probability that two white balls are
chosen let's figure out the probabilities by multiplying probabilities along the
branches 3 over 7 * 2 over 6 = 6 over 42 3/ 7 * 4/ 6 = 12/
42 4/ 7 * 3/ 6 = 12/ 42 4 over 7 * 3/ 6 = 12 over 42 then we need to label white
white white red red white red
red we are looking for two white balls which means this one so 6 over
42 is the solution back B contains four white balls and three red balls when two balls
are chosen at random without replacement from back B the probability that they are both white is 2 over
seven a standard die is rolled if one or two is obtained two balls are chosen without replacement from back a
otherwise they are chosen from back B based on this information we need to create a tree
diagram from this uh root note back a go here back
B we do know the probability of choosing two white balls is 2 over
s white white for back B one or two is a obtain the two balls are chosen without replacement from back
a 2 over 6 4 over 6 for choosing back B find the
probability that the two balls are white based on this uh new tree diagram
first of all you need to roll a DI then you decide it's a back a or back B for back a the probability of choosing
two white balls is 6 over 42 the probability of choosing two white balls is a 2 over 7 we need to multiply
by 4 over 6 for this 2 over 7 this is the new tree diagram for two white balls for back
eight two over 6 time 6 over 42 work this out 1 over 21 this
equals 4 over 21 therefore for two white we need to add
these two together 1 over 21 + 4 over 21 = 5 over 21 B given that both balls are white
that means 5 over 21 go to denominator find the probability that
they were chosen from B A this is the probability for B A
therefore 1 over 21 over 5 over 21 = 1 over 5 is the answer
Heads up!
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