Probability - Aptitude MCQ Questions and Solutions with Explanations

At least one tails means one tails or both tails
Let us see what the possible outcomes are when two coins are tossed simultaneously
We can get,
1. Heads and Heads
2. Heads and Tails
3. Tails and Heads
4. Tails and Tails
So we have total = 4 outcomes
Out of them, 3 have tails in it
Chances of at least 1 tails = $$\frac{{3}}{{4}}$$

From box 1 we can pick black or green shirt
Case 1: Pick black shirt
Box 1 has total 5 + 3 = 8 shirts
Probability of black from box 1 = $$\frac{{5}}{{8}}$$
Now this black is added to box 2
So box 2 now has 3 + 1 = 4 black & 5 green shirts
Total = 4 + 5 = 9 shirts
Probability of black from box 2 = $$\frac{{4}}{{9}}$$
Case 1 probability
$$\eqalign{ & = \frac{5}{8} \times \frac{4}{9} \cr & = \frac{{20}}{{72}} \cr} $$

Case 2: Pick green shirt
Probability of green from box 1 = $$\frac{{3}}{{8}}$$
Now this green is added to box 2
So box 2 now has 3 black & 5 + 1 = 6 green shirts
Total = 3 + 6 = 9 shirts
Probability of black from box 2 = $$\frac{{3}}{{9}}$$
Case 2 probability
$$\eqalign{ & = \frac{3}{8} \times \frac{3}{9} \cr & = \frac{9}{{72}} \cr} $$
Total Probability
$$\eqalign{ & = \frac{{20}}{{72}} + \frac{9}{{72}} \cr & = \frac{{29}}{{72}} \cr} $$

We want green card or IMPORTANT card
There are 30 - 17 = 13 green cards
There are 4 + 5 = 9 IMPORTANT cards
Total cards = 30
Also 5 green cards are IMPORTANT cards
So probability
$$\eqalign{ & = \frac{{13}}{{30}} + \frac{9}{{30}} - \frac{5}{{30}} \cr & = \frac{{17}}{{30}} \cr} $$

Since a pair has 2 socks in it and there are just two colors - white and grey, when Suresh chooses 3 socks, he will surely select 2 socks of same color - be it grey or white.
So possibility = probability = 1

Total socks = 4 + 3 + 5 = 12
We want same color socks
So we want, 2 white or 2 blue or 2 grey socks
For white :
Probability of 1^st sock being white = $$\frac{{4}}{{12}}$$
Probability of 2^nd sock being white = $$\frac{{3}}{{11}}$$
White Probability
$$\eqalign{ & = \frac{4}{{12}} \times \frac{3}{{11}} \cr & = \frac{1}{{11}} \cr} $$
Similarly,
Blue Probability
$$\eqalign{ & = \frac{3}{{12}} \times \frac{2}{{11}} \cr & = \frac{1}{{22}} \cr} $$
Grey Probability
$$\eqalign{ & = \frac{5}{{12}} \times \frac{4}{{11}} \cr & = \frac{5}{{33}} \cr} $$
∴ Total Probability
$$\eqalign{ & = \frac{1}{{11}} + \frac{1}{{22}} + \frac{5}{{33}} \cr & = \frac{{19}}{{66}} \cr} $$

We want 2 red and 1 blue shirt
There are 4 red shirts and 3 blue shirts
Total = 15 shirts
You can choose blue shirt 1^st, then red shirt
And then red shirt Probability
$$\eqalign{ & = \frac{3}{{15}} \times \frac{4}{{14}} \times \frac{3}{{13}} \cr & = \frac{6}{{455}} \cr} $$
Or you can choose red shirt 1^st, then red shirt and then blue shirt
Or you can choose red shirt 1^st, then blue shirt and then red shirt
For all 3 the probability remains same = $$\frac{6}{{455}}$$
We need to add these 3 probabilities to get total probability
∴ Total probability
$$\eqalign{ & = \frac{6}{{455}} + \frac{6}{{455}} + \frac{6}{{455}} \cr & = \frac{{18}}{{455}} \cr} $$

We want black or green ball
There are 6 black and 8 green balls
Total balls = 2 + 6 + 4 + 8 = 20
So Probability $$ = \frac{6}{{20}} + \frac{8}{{20}} = \frac{7}{{10}}$$

Number of Black socks = 5
Total socks = 5 + 3 = 8
First we draw one sock
Probability of it being black = $$\frac{{5}}{{8}}$$
Now a second sock is picked
Probability of its being black = $$\frac{{4}}{{7}}$$
Total Probability $$ = \frac{5}{8} \times \frac{4}{7} = \frac{5}{{14}}$$

Choosing 4 shirts means 1^st shirt and 2^nd and then 3^rd and then 4^th shirt
At least 1 red shirt means there can be 1, 2, 3 or 4 red shirts
So, Probability of choosing red = 1 - Probability of not choosing red
If we remove red shirts, then 15 - 4 red shirts = 11 shirts remain.
We have to choose 4 out of these 11
So Probability of choosing 4 shirts which are not red
$$\eqalign{ & = \frac{{11}}{{15}} \times \frac{{10}}{{14}} \times \frac{9}{{13}} \times \frac{8}{{12}} \cr & = \frac{{22}}{{91}} \cr} $$
∴ Probability of picking at least 1 red shirt
$$\eqalign{ & = 1 - \frac{{22}}{{91}} \cr & = \frac{{69}}{{91}} \cr} $$

Both white or both blue
Total shirts = 15
There are 2 white and 3 blue shirts
Probability for 2 white shirts $$ = \frac{2}{{15}} \times \frac{1}{{14}} = \frac{1}{{105}}$$
Probability for 2 blue shirts $$ = \frac{3}{{15}} \times \frac{2}{{14}} = \frac{1}{{35}}$$
Total Probability $$ = \frac{1}{{105}} + \frac{1}{{35}} = \frac{4}{{105}}$$