CK-12-Basic Probability and Statistics Concepts - A Full Course

4.4. Binomcdf Function http://www.ck12.org

P(X=a) =nCa×pa×q(n−a)

P(2 girls) = 5 C 2 ×p^2 ×q^3

P(2 girls) = 5 C 2 ×( 0. 50 )^2 ×( 0. 50 )^3

P(2 girls) = 10 × 0. 25 × 0. 125

P(2 girls) = 0. 3125

P(1 girl) = 5 C 1 ×p^1 ×q^4

P(1 girl) = 5 C 1 ×( 0. 50 )^1 ×( 0. 50 )^4

P(1 girl) = 5 × 0. 50 × 0. 0625

P(1 girl) = 0. 1563

P(0 girls) = 5 C 0 ×p^0 ×q^5

P(0 girls) = 5 C 0 ×( 0. 50 )^0 ×( 0. 50 )^5

P(0 girls) = 1 × 1 × 0. 03125

P(0 girls) = 0. 03125

The total probability for this example is calculated as follows:

P(X< 3 ) = 0. 3125 + 0. 1563 + 0. 3125

P(X< 3 ) = 0. 500

Therefore, the probability of havingless than3 girls in 5 children is 50.0%.

When using technology, you will select the binomcdf function, because you are looking for the probability ofless
than3 girls from the 5 children.

Example C

A fair coin is tossed 50 times. What is the probability that you will get heads inat most30 of these tosses?

There are 50 trials, son=50.

A success is getting a head, and we are interested inat most30 heads. Therefore,a= 30 , 29 , 28 , 27 , 26 , 25 , 24 , 23 , 22 , 21 , 20 , 19 , 18 , 17 , 16 , 15 , 14 , 13 , 12 , 11 , 10 , 9 , 8 , 7 , 6 , 5 , 4 , 3 , 2 ,1,
and, 0.