six vitamins and three sugar tablets are in a box and they a

six vitamins and three sugar tablets are in a box and they all look the same. One tablet is given at random and given to person A. Then a tablet is selected and given to person B. What is the probability that: 1. Person A was given a vitamin. 2. person B was given a sugar tablet given that person A was given a vitamin. 3. neither was given a vitamin. 4. both was given a vitamin. 5. both was given a sugar tablet?

Solution

a)

From 9 tablets, 6 are vitamins

So, probability that A got a Vitamin tablet = 6 / 9 = 2/3

b)

Here we know that person A was given a vitamin. Now there are 8 tablets left out of which 3 are sugars.

Thus, required probability = 3 / 8

c)

Probability of neither being given a vitamin would be when both of them are give n sugars,

Prob. of A getting sugar = 3 / 9

Prob. of B getting sugar = 2 / 8

Thus, both occuring simultaneously = 3/9 * 2/8 = 6 / 72 = 1 / 12

d)

Prob. of A getting a vitamin = 6 /9

Prob. of B getting a Vitamin = 5 / 8

Both happening together = 6/9 * 5/8 = 30 / 72 = 5 / 12

e)

Probabilty of both being given a sugar tablet = probabiliy of neither getting a vitamin

Therefore, same answer as part c

Probability = 1 / 12

Hope this helps. Ask if you have any doubts.