really confused..please help? picture attached:)

Question

anonymous · Answer

attachment

anonymous · Answer

@amistre64

anonymous · Answer

@ganeshie8

ganeshie8 · Answer

Is this question from probability  ?

anonymous · Answer

i think so..?

ganeshie8 · Answer

$$ (p+q)^n =   \sum \limits_{k=1}^n  ~^nC_k p^{n-k}q^{k}$$

ganeshie8 · Answer

Oh, then im not sure of the binomial distribution in probability @amistre64

ranga · Answer

If p =0.3, q = 1 - p = 1 - 0.3 = 0.7
n = 5
Plug in the numbers above.

anonymous · Answer

what do I do after i plug in the numbers?

ranga · Answer

The Probability of getting EXACTLY "r" successes, P(r), out of "n" trails is:
$$P(r) = ~^nC_rp^{r}q^{n-r}~;~~~~n = 5; ~~p = 0.3; ~~q = 0.7\P(r) = ~^{5}C_r(0.3)^{r}(0.7)^{5-r}\P(0) = ~^{5}C_0(0.3)^{0}(0.7)^{5-0} = ~?\P(1) = ~^{5}C_1(0.3)^{1}(0.7)^{5-1} =~ ?\P(2) = ~^{5}C_2(0.3)^{2}(0.7)^{5-2} =~ ?\P(3) = ~^{5}C_3(0.3)^{3}(0.7)^{5-3} =~ ?\P(4) = ~^{5}C_4(0.3)^{4}(0.7)^{5-4} =~ ?\P(5) = ~^{5}C_5(0.3)^{5}(0.7)^{5-5} =~ ?$$That will be the binomial distribution for the given data.