A bag contains x red balls, 12 blue balls, and 8 green balls. The theoretical probability … — JAMB UTME

A bag contains $x$ red balls, $12$ blue balls, and $8$ green balls. The theoretical probability of selecting a green ball at random from the bag is $\frac{1}{4}$ . In an experiment where a ball is drawn with replacement $240$ times, a blue ball is recorded $102$ times. What is the positive difference between the experimental relative frequency and the theoretical probability of drawing a blue ball?

$\frac{1}{20}$ Cevap
B
$\frac{7}{40}$
C
$\frac{3}{8}$
D
$\frac{17}{40}$

Cevap

The positive difference between the experimental relative frequency and the theoretical probability of drawing a blue ball is

\frac{1}{20}

The total number of balls in the container is determined from the theoretical probability of green balls:

\frac{8}{N} = \frac{1}{4} \implies N = 32

. The theoretical probability of drawing a blue ball is

\frac{12}{32} = \frac{3}{8} = \frac{15}{40}

. The experimental relative frequency is

\frac{102}{240} = \frac{17}{40}

. Taking the positive difference yields

\frac{17}{40} - \frac{15}{40} = \frac{2}{40} = \frac{1}{20}

Adım Adım Çözüm

Calculate total number of balls in the bag

Total balls

N = 32

Given theoretical

P(\text{Green}) = \frac{8}{N} = \frac{1}{4}

, solving for

N

yields

N = 32

Determine theoretical probability of selecting a blue ball

P(\text{Blue}) = \frac{3}{8} = \frac{15}{40}

There are

12

blue balls out of

32

total balls, so

P(\text{Blue}) = \frac{12}{32} = \frac{3}{8}

Calculate experimental relative frequency of selecting a blue ball

Relative frequency =

\frac{17}{40}

Blue was drawn

102

times in

240

trials, so

\frac{102}{240} = \frac{17}{40}

Find the positive difference between experimental relative frequency and theoretical probability

Difference =

\frac{1}{20}

Subtracting

\frac{15}{40}

from

\frac{17}{40}

gives

\frac{2}{40} = \frac{1}{20}

Anahtar Kavram

Experimental relative frequency vs theoretical probability comparison

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