Question

Out of 100 balls in a bag, 25 are red, 30 are blue and 45 are white. Find the probability that a ball drawn from the bag is red; (i) blue; (ii) white; (iv) Find the sum of the probabilities got in (i), (ii) and (iii).​

Answer 1

Answer:

(i)red=25/100=1/4

(ii)blue=30/100=3/10

(iii)white=45/100=9/20

(iv)red ball+blue ball+white ball=1/4+3/10+9/20

=(1×5+3×2+9×1)/20

=(5+6+9)/20

=20/20

=1

Answer 2

p(R)=1/4, p(B)=3/10, p(W)=9/20

Sum of probabilities of all three events is 1.

i.e. $\bf p(R) +p(B)+p(W)= 1$

Given:

• Out of 100 balls in a bag,25 are red,
• 30 are blue and 45 are white.

To Find:

• Find the probability that a ball drawn from the bag is
• (i) red;
• (ii) blue;
• (iii) white;
• (iv) Find the sum of the probabilities got in (i), (ii) and (iii).

Solution:

Formula to be used:

Probability= favourable outcomes/Total outcomes.

Step 1:

Find the probability of getting red ball.

Total balls: 100

Red balls: 25

Let the probability of red ball is p(R), so

$p(R) = \frac{25}{100}$

or

$\bf p(R) = \frac{1}{4} ...eq1$

Step 2:

Find the probability of getting blue ball.

Total balls: 100

Red balls: 30

Let the probability of red ball is p(B), so

$p(B) = \frac{30}{100}$

or

$\bf p(B) = \frac{3}{10} ...eq2$

Step 3:

Find the probability of getting white ball.

Total balls: 100

Red balls: 45

Let the probability of red ball is p(W), so

$p(W) = \frac{45}{100}$

or

$\bf p(W) = \frac{9}{20} ...eq3$

Step 4:

Find the sum of probabilities of step 1,2 and 3.

$p(R) +p(B)+p(W)= \frac{25}{100} + \frac{30}{100} + \frac{45}{100}$

or

$= \frac{25 + 30 + 45}{100}$

or

$= \frac{100}{100}$

or

$\bf p(R) +p(B)+p(W)= 1$

Thus,

p(R)=1/4, p(B)=3/10, p(W)=9/20

Sum of probabilities of all three events is 1.

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