CRE 6 - Conditional Probability | Modern Math - Probability

Answer: Option C

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Explanation :

When two die are rolled total possible outcomes = 6 × 6 = 36.

Out of these 36 outcomes, 6 outcomes result in same number on both dice.

Since the number on two dice are different, possible outcomes here = 36 – 6 = 30.

∴ Desired outcomes for sum to be 6 = {(1, 5) or (5, 1) or (2, 4) or (4, 2)} i.e., 4 outcomes.

∴ Required probability = 4/30 = 2/15

Hence, option (c).

Answer: Option B

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Explanation :

There are 11 integers: 6 odd & 5 even.

Sum of two integers is even when either both are even or both are odd.

Both even: Number of ways of selecting two even integers = ⁵C₂ = 10.

Both odd: Number of ways of selecting two odd integers = ⁶C₂ = 15.

∴ Total possible outcomes = 10 + 15 = 25.

∴ Desired outcomes for both integers to be odd = 15.

∴ Required probability = 15/25 = 3/5

Hence, option (b).

Answer: Option A

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Explanation :

Total possible outcomes for sum to be 6 = {(1, 5) or (5, 1) or (2, 4) or (4, 2) or (3, 3)} i.e., 5 outcomes.

Here, desired outcomes = {(2, 4) or (4, 2)} i.e., 2 outcomes.

∴ Required probability = 2/5

Hence, option (a).

Answer: Option C

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Explanation :

Total possible outcomes when both cards are red = ²⁶C₂ = 325.

Now, there are 18 red number cards available.

∴ Desired outcomes for both the cards to have different numbers = ¹⁸C₂ – 9 = 153 – 9 = 144

[out of total ¹⁸C₂ ways of selecting number cards 9 ways will give us same number on both cards.]

∴ Required probability = 144/325

Hence, option (c).

Answer: Option A

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Explanation :

Let us calculate P (selecting 1^st urn and drawing a red ball from it).

Out of three urns probability of selecting 1^st urn = 1/3.

P (drawing a red ball from 1^st urn) = $\frac{6}{10}$ = $\frac{3}{5}$

P (selecting 1^st urn and drawing a red ball from it) = $\frac{1}{3}\times \frac{3}{5}$ = $\frac{3}{15}$

Similarly,

P (selecting 2^nd urn and drawing a red ball from it) = $\frac{1}{3}\times \frac{2}{5}$ = $\frac{2}{15}$

P (selecting 3^rd urn and drawing a red ball from it) = $\frac{1}{3}\times \frac{1}{2}$ = $\frac{1}{6}$

P (If the ball drawn is red, find the probability that it is drawn from the first urn) 

= $\frac{\mathrm{P}\left(\mathrm{red}\mathrm{ }\mathrm{ball}\mathrm{ }\mathrm{from}\mathrm{ }1\mathrm{st}\mathrm{ }\mathrm{urn}\right)}{\mathrm{P}\left(\mathrm{red}\mathrm{ }\mathrm{ball}\mathrm{ }\mathrm{from}\mathrm{ }1\mathrm{st}\mathrm{ }\mathrm{urn}\right)+\mathrm{P}\left(\mathrm{red}\mathrm{ }\mathrm{ball}\mathrm{ }\mathrm{from}\mathrm{ }2\mathrm{nd}\mathrm{ }\mathrm{urn}\right)+\mathrm{P}\left(\mathrm{red}\mathrm{ }\mathrm{ball}\mathrm{ }\mathrm{from}\mathrm{ }3\mathrm{rd}\mathrm{ }\mathrm{urn}\right)}$ 

= $\frac{\frac{3}{15}}{\frac{3}{15}+\frac{2}{15}+\frac{1}{6}}$ = $\frac{\frac{3}{15}}{\frac{6+4+5}{30}}$ = $\frac{3}{15}\times \frac{30}{15}$ = $\frac{2}{5}$

Hence, option (a).