Level 6 · Ages 16–17
Read it again any time, including while you work through the exercises.
Some experiments have only two outcomes: success or failure. Tossing a coin and hoping for heads, drawing a card and hoping for a heart, attempting a basket. Such an experiment is called a Bernoulli trial, with a success probability written p and a failure probability 1 - p.
Now repeat this trial n times, independently and always under the same conditions, and count the number of successes. This number, call it X, follows what is known as the binomial distribution with parameters n and p. The whole question becomes: what is the probability of getting exactly k successes?
The tree gives the answer. Each path with k successes and n - k failures has probability p to the power k times (1 - p) to the power (n - k): multiply along the path. But several paths give the same number of successes! For 1 success in 2 tosses: success-failure and failure-success, two paths. So you must MULTIPLY the probability of one path by the number of paths.
Hence the formula: P(X = k) = (number of paths with k successes) × p to the k × (1 - p) to the (n - k). This number of paths is written "n choose k": the binomial coefficient. For small values, the tree is enough to count it: 3 choose 2 is 3, 4 choose 2 is 6.
One last treasure: the expected value. On average, over n trials, you get n × p successes. Twenty shots with a 30% success rate? Expect about 20 × 0.3 = 6 baskets. Simple, powerful, and you will see that it works.
We toss a fair coin twice and count the number of heads, written X. Compute P(X = 1).
chemins à 1 pile : PF et FP -> 2 chemins
On the tree, I look for the paths with exactly one heads: heads then tails, and tails then heads. There are 2.
1 chemin : 0,5 × 0,5 = 0,25
Each path has probability 0.5 × 0.5 = 0.25: multiply along the path.
P(X = 1) = 2 × 0,25 = 0,5
The 2 paths give the same event, so I add their probabilities: 2 times 0.25 equals 0.5.
Result: P(X = 1) = 0.5.
An archer hits the target with probability 0.2 on each shot. She shoots 3 arrows, independently. X counts the successful shots. Compute P(X = 1).
1 chemin : 0,2 × 0,8 × 0,8 = 0,128
A path with 1 success and 2 failures has probability 0.2 × 0.8 × 0.8 = 0.128.
3 chemins : SEE, ESE, EES
The success can happen on the 1st, 2nd or 3rd shot: 3 paths lead to exactly 1 success.
P(X = 1) = 3 × 0,128 = 0,384
I multiply one path's probability by the number of paths: 3 × 0.128 = 0.384.
Result: P(X = 1) = 0.384.