Question

A box contains 4 chocobar and 4 ice creams.Tom eats 3 of them by randomly choosing find the probablity of choosing 2 chocolate and 1 ice cream solution

Answer 1

sample space(total)=no.of chocobar+no. of ice cream

=4+4

=8

P(2chocolate and 1ice cream)=3/8

hope it helps...

Answer 2

Answer:

Given:

Probability of taking chocobar from packet of 8 =4/8 =0.5 =P(A)

Probability of taking ice cream from packet of 8 =4/8  =0.5  =P(B)

Step-by-step explanation:

Tom eat 3 of them in random choosing,

Ways in which 2 chocobars and 1 ice cream can be chosen=

AAB

ABA

BAA

So probability of choosing 2 chocobars and 1 ice creams=3*0.5*0.5*0.5=3/8

(when considering with replacement  which is not applicable here)

Here, if not replaced Probability for above case = 3*(4/8)*(3/7)*(4/6)= 3/7

(∵4 chocobar from packet of 8 * 3 chocobar from packet of 7(as one taken) * 4 ice cream from rest 6 in packet  * 3(total combinations)

Regards,

Chengizkhan