Question

A Bag contains 3 red apples and 4 green apples. An apple is drawn from the bag without looking into the bag. What is the probability of getting a green apple?​

Answer 1

Step-by-step explanation:

e, event of getting a green apple

p(e) = no. of green apples/total no. of apples

= 4/(4+3)=4/7

4/7 is probability of getting a green Apple

Answer 2

★GIVEN★

• A Bag contains 3 red apples and 4 green apples.
• An apple is drawn from the bag without looking into the bag.

★To Find★

Probability of getting a green apple.

★SOLUTION★

We know that formula of probability is,

$\large{\green{\underline{\boxed{\bf{P(E)=\dfrac{No\:of\:favourable\:outcomes}{Total\:no\:of\:possible\:outcomes}}}}}}$

Total no of possible outcomes = 3 + 4 = 7.

Now,

Favourable outcomes for green apple = 4.

According to the question,

$\large\implies{\sf{P_{(green\:apple)}=\dfrac{No\:of\:favourable\:outcome}{Total\:no\:of\:possible\:outcomes}}}$

$\large\therefore\boxed{\bf{P_{(green\:apple)}=\dfrac{4}{7}}}$

$\large{\green{\underline{\boxed{\therefore{\bf{Probability\:of\:getting\:a\:green\:apple\:is\:\dfrac{4}{7}.}}}}}}$