Here’s how it’s done: you count the total number of m&ms to start.
then, you count the number of each color of m&ms. i’m just making up numbers right now but say there are 50 m&ms total and 10 of them are blue. you would put the number of blue m&ms over the total number of m&ms. this would make the probability you pull a blue m&m 10/50. you would then simplify the answer to 1/5. that would be what u put in the box. remember, i just made up those numbers so that’s not the real answer but hopefully u get the point haha. if ur still confused ask me

3 0

3 years ago

"The manager for State Bank and Trust has recently examined the credit card account balances for the customers of her bank and f

yuradex [85]

Answer:

$P(X = 4 ) = 0.1876$

$P(X \le 4) = 0.8358$

Step-by-step explanation:

From the question we are told that

The proportion that has outstanding balance is p = 0.20

The sample size is n = 15

Given that the properties of the binomial distribution apply, for a randomly selected number(X) of credit card

$X \ \ ~ Bin (n , p )$

Generally the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards is mathematically represented as

$P(X = 4 ) = ^nC_4 * p^4 * (1 - p)^{n-4}$

=> $P(X = 4 ) = ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}$

Here C stand for combination

=> $P(X = 4 ) = 0.1876$

Generally the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card is mathematically represented as

$P(X \le 4) = [ ^{15}C_0 * (0.20)^0 * (1 - 0.20)^{15-0}]+[ ^{15}C_1 * (0.20)^1 * (1 - 0.20)^{15-1}]+\cdots+[ ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}]$

=> $P(X \le 4) = 0.8358$

4 0

3 years ago

B.48(a’b2c2 + a²b3c2 + a²b²c3)- 2a2b2c2​

B.48(a’b2c2 + a²b3c2 + a²b²c3)- 2a2b2c2