Question

@ Moderators @ Stars @ Best Users # Question # Two groups are competing for the position on boards of directors of a corporation . the probability that the first group and the second group will win are 0.6 and 0.4 respectively. father is the first group wins the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins . find the probability that the new product introduced was bye second group .​

Answer 1

let

Consider F as first group, S as second group and N as introducing a new product

Here we should determine

P (S/N) – New product introduced by the second group

$p \: by \: s \: = \frac{p(s).p(n \: and \: s)}{p(s).p(n \: and \: s) + p(f).p(n \: and \: s)}$

substitute the value

$= \frac{(0.4)(0.3)}{(0.6)(0.7) + (0.4)(0.3)}$

so we get

=0.12/0.54

=2/9

hence the probability of the second indrotused a new product is 2/9

Answer 2

Answer:

$@Moderators\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ @ Stars \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ @Best Users \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ \begin{gathered} \bf {Let \: I_n = \int { tan}^{n} \: x \: dx(n > 1)} \\ \\ \bf If \: I_4 + I_6 = a. {tan}^{5}x + b {x}^{5} + C\end{gathered}LetIn=∫tannxdx(n>1)IfI4+I6=a.tan5x+bx5+C \ \textless \ br /\ \textgreater \ Then Find Values of a & b \ \textless \ br /\ \textgreater$

@Moderators

@ Stars

@Best Users

\begin{gathered} \bf {Let \: I_n = \int { tan}^{n} \: x \: dx(n > 1)} \\ \\ \bf If \: I_4 + I_6 = a. {tan}^{5}x + b {x}^{5} + C\end{gathered}

LetI

n

=∫tan

n

xdx(n>1)

IfI

4

+I

6

=a.tan

5

x+bx

5

+C

Then Find Values of a & b