Coin has a probability of 0.7 of coming up heads and 0.3 of tails. if thrown 5 times.

whats the probability of getting
1) exactly 2 heads
b) 5 tails
c) exactly 1 tail
d) 5 heads
e) exactly 1 head
thanks in advance

Let X = number of heads in 5 flips of the unbalanced coin.

a) P(X = 2) = 5C2 * .7^2 * .3^3 = .1323
b) P(X = 0) = 5C0 * .7^0 * .3^5 = .00243 (5 tails is the same as 0 heads)
c) P(X = 4) = 5C4 * .7^4 * .3^1 = .36015 (1 tail is the same as 4 heads)
d) P(X = 5) = 5C5 * .7^5 * .3^0 = .16807
e) P(X = 1) = 5C1 * .7^1 * .3^4 = .02835

a) (.7)(.7)(.3)(.3)(.3) =.01323
b) (.3)(.3)(.3)(.3)(.3) = .00243
c) (.3)(.7)(.7)(.7)(.7) = .07203
d) (.7)(.7)(.7)(.7)(.7) = .016807
e) (.7)(.3)(.3)(.3)(.3) = .00567

X = number of heads in 5 tosses
X ~ Bi(5 , 0.7)

use the formula to answer a/...e/