Hello
I have attached the question becuase it wasn't in word format.
I need help with part iii)
but my answers to first two parts are :
i)4/10*3/9*2/8*1/7=1/210
ii)3/10*2/9*4/8*3/7=1/70
however i didn't understand part iii)
thanks
Hello
I have attached the question becuase it wasn't in word format.
I need help with part iii)
but my answers to first two parts are :
i)4/10*3/9*2/8*1/7=1/210
ii)3/10*2/9*4/8*3/7=1/70
however i didn't understand part iii)
thanks
Because of the limited numbers here the third part is best done head on.
Find the probability of two reds, one orange and one blue.
Find the probability of one red, two oranges and one blue.
Find the probability of one red, one orange and two blues.
Then add these up to have at least one of each.
Hi master_2m8,
Think your part ii) is interpreted wrongly, question request for 2 orange and 2 blue in any order, however 1/70 is the answer for choosing 2 orange 1st before choosing 2 blue.
$\displaystyle P(\text{2 orange and 2 blue})= \frac{{3 \choose 2}.{4 \choose 2}}{{10 \choose 4}}=\frac{3}{35}$