A bag contains 2 yellow balls, 3 green balls, 5 red balls and 6 black balls.

Mathematics

2 answers:

shepuryov [24]2 years ago

5 0

Answer:

7/16

Step-by-step explanation:

(2+5)/(2+3+5+6) = 7/16

natali 33 [55]2 years ago

3 0

The probability of either a yellow or a red ball being drawn if only one is drawn is 7/16.

<h3>What is probability?</h3>

The probability is the measure of the likelihood of an event to happen. It measures the certainty of the event. The formula for probability is given by; P(E) = Number of Favourable Outcomes/Number of total outcomes.

According to the given question.

Numnber of yellow balls = 2

Number of green balls = 3

Number of red balls = 5

Number of black balls = 6

So,

The total number of balls in a bag = 2+ 3+ 5 + 6 = 16

And,

The sum of yellow and red ball = 2 + 5 = 7

Therefore,

The probability of either a yellow or a red ball being drawn if only one ball is drawn

= $\frac{total\ number\ of \ red \ and \ yellow\ balls }{Total\ number\ of \ balls}$

$= \frac{7}{16}$

Hence, the probability of either a yellow or a red ball being drawn if only one is drawn is 7/16.

Find out more inforation about probability here:

brainly.com/question/11234923

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Rose and Jack plan to study together for the Math test. They decide to meet at the library between 8:00pm and 8:30pm. Assume tha

natulia [17]

Answer:

(a) 1/3

(b) 1/15

Step-by-step explanation:

(a)Let X denote the waiting time in minutes. It is given that X follows a uniform distribution, and since the variable being measured is time, we assume it to be a continuous uniform distribution.

$f_X(x) =\begin{cases} \frac{1}{30} & 0\leqx\leq 30\\ 0 & otherwise \end{cases}$

Now

$P(X>20) = \int_{20}^{30}f_X(x) = \frac{1}{30} \times 10 = \frac{1}{3}$ .

Jack or Rose arriving first is equally likely, therefore the probability of Jack waiting is just the half of the above obtained probability i.e $\frac{1}{6}$