There are "n" delicious sweets in the bag. You know that 6 of them are orange and the rest are yellow. Hannah takes a random swe
et from the bag and eats it. Then she takes a second sweet from the bag and eats that one. You know that the probability that Hannah eats two of the orange sweets is 1/3. How many sweets were in the bag before Hannah ate any
At the point when Hannah takes her first sweet from the sack, there is a 6/n chance it is orange. This is because that there are 6 orange desserts and n desserts altogether. When Hannah takes out her second sweet, there is a 5/(n-1) chance that it is orange. This is because there are just 5 orange desserts let alone for an aggregate of n-1 desserts. The possibility of getting two orange desserts in succession is the main likelihood increased by the second one: 6/n x 5/n–1 The question lets us know that the shot of Hannah getting two orange desserts is 1/3. So: 6/n x 5/n–1 = 1/3 Now, rearrange this problem. (6x5)/n(n-1) = 1/3 This gets to be: 30/(n² – n) = 1/3 Times by 3 on both sides: 90/(n² – n) = 1 What's more, doing likewise with (n² – n): So (n² – n) = 90 Our answer is: n² – n – 90 = 0
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