Using the Probability Mass Function Over a Range

Other
1

Lesson: https://www.codecademy.com/journeys/data-scientist-ml/paths/dsmlcj-22-data-science-foundations-ii/tracks/dsmlcj-22-statistics-fundamentals-for-data-science/modules/dsf-probability-for-data-science-530e3dd3-f0d5-4e18-8106-666e62adf717/lessons/introduction-to-probability-distributions/exercises/using-the-probability-mass-function-over-a-range

How do i calculate P(X=1)+P(X=2)+P(X=3) individually to get the results 0.1562+0.3125+0.3125 ? Could be pretty simple but im kinda stuck

No one answered user bonoboboc, so I thought I’d create a new thread.
Forums: FAQ: Introduction to Probability Distributions - Using the Probability Mass Function Over a Range

2

We are tossing a coin 5 times.

There are a total of 25 or 32 outcomes (e.g. if T denotes a tail and H denotes a head, then the sample space will have 32 possibilities: TTTTT, HTTTT, THTTT, ..., THHHH, HHHHH)

0 heads —>   5C0 —> 1 outcome

Probability of 0 heads = 1⁄32 = 0.03125

1 heads —>   5C1 —> 5 outcomes

Probability of 1 heads = 5⁄32 = 0.15625

2 heads —>   5C2 —> 10 outcomes

Probability of 2 heads = 10⁄32 = 0.3125

3 heads —>   5C3 —> 10 outcomes

Probability of 3 heads = 10⁄32 = 0.3125

4 heads —>   5C4 —> 5 outcomes

Probability of 4 heads = 5⁄32 = 0.15625

5 heads —>   5C5 —> 1 outcome

Probability of 5 heads = 1⁄32 = 0.03125

(Check)
Sum of probabilities of all outcomes = 0.03125 + 0.15625 + 0.3125 + 0.3125 + 0.15625 + 0.03125 = 1

Probability of getting 1 to 3 heads = (Probability of 1 head) + (Probability of 2 heads) + (Probability of 3 heads) = 0.15625 + 0.3125 + 0.3125 = 0.78125

scrn1
scrn1690×564 6.67 KB